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Course: Algebra 1   >   Unit 14

  • Finding the vertex of a parabola in standard form
  • Graphing quadratics: standard form
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Quadratic word problem: ball

  • Quadratic word problems (standard form)

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How to Solve Word Problems Requiring Quadratic Equations

Last Updated: December 27, 2020

wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, volunteer authors worked to edit and improve it over time. This article has been viewed 14,258 times.

Some word problems require quadratic equations in order to be solved. In this article, you will learn how to solve those types of problems. Once you get the hang of it, it will be very easy.

Quadratic Equations

Step 1 Know what kind of problem you're tackling.

  • For the real life scenarios, factoring method is better.
  • In geometric problems, it is good to use the quadratic formula.

Real Life Scenario

Step 1 Ask to yourself,

  • In this problem, it asks for Kenny's birthday.

Step 2 Decide your variables.

  • Since negative month does not exist, 3 is the only one that makes sense.
  • Because the problem asks for both the month and the date, the answer would be March 18th. (Use the value for the other variable that you found in step 3.)

Geometric Problems

Step 1 Identify if it's a geometric problem.

  • In the problem above, it asks you only for the height of the triangle.

Step 3 Decide your variables.

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Chapter 10: Quadratics

10.7 Quadratic Word Problems: Age and Numbers

Quadratic-based word problems are the third type of word problems covered in MATQ 1099, with the first being linear equations of one variable and the second linear equations of two or more variables. Quadratic equations can be used in the same types of word problems as you encountered before, except that, in working through the given data, you will end up constructing a quadratic equation. To find the solution, you will be required to either factor the quadratic equation or use substitution.

Example 10.7.1

The sum of two numbers is 18, and the product of these two numbers is 56. What are the numbers?

First, we know two things:

[latex]\begin{array}{l} \text{smaller }(S)+\text{larger }(L)=18\Rightarrow L=18-S \\ \\ S\times L=56 \end{array}[/latex]

Substituting [latex]18-S[/latex] for [latex]L[/latex] in the second equation gives:

[latex]S(18-S)=56[/latex]

Multiplying this out gives:

[latex]18S-S^2=56[/latex]

Which rearranges to:

[latex]S^2-18S+56=0[/latex]

Second, factor this quadratic to get our solution:

[latex]\begin{array}{rrrrrrl} S^2&-&18S&+&56&=&0 \\ (S&-&4)(S&-&14)&=&0 \\ \\ &&&&S&=&4, 14 \end{array}[/latex]

[latex]\begin{array}{l} S=4, L=18-4=14 \\ \\ S=14, L=18-14=4 \text{ (this solution is rejected)} \end{array}[/latex]

Example 10.7.2

The difference of the squares of two consecutive even integers is 68. What are these numbers?

The variables used for two consecutive integers (either odd or even) is [latex]x[/latex] and [latex]x + 2[/latex]. The equation to use for this problem is [latex](x + 2)^2 - (x)^2 = 68[/latex]. Simplifying this yields:

[latex]\begin{array}{rrrrrrrrr} &&(x&+&2)^2&-&(x)^2&=&68 \\ x^2&+&4x&+&4&-&x^2&=&68 \\ &&&&4x&+&4&=&68 \\ &&&&&-&4&&-4 \\ \hline &&&&&&\dfrac{4x}{4}&=&\dfrac{64}{4} \\ \\ &&&&&&x&=&16 \end{array}[/latex]

This means that the two integers are 16 and 18.

Example 10.7.3

The product of the ages of Sally and Joey now is 175 more than the product of their ages 5 years prior. If Sally is 20 years older than Joey, what are their current ages?

The equations are:

[latex]\begin{array}{rrl} (S)(J)&=&175+(S-5)(J-5) \\ S&=&J+20 \end{array}[/latex]

Substituting for S gives us:

[latex]\begin{array}{rrrrrrrrcrr} (J&+&20)(J)&=&175&+&(J&+&20-5)(J&-&5) \\ J^2&+&20J&=&175&+&(J&+&15)(J&-&5) \\ J^2&+&20J&=&175&+&J^2&+&10J&-&75 \\ -J^2&-&10J&&&-&J^2&-&10J&& \\ \hline &&\dfrac{10J}{10}&=&\dfrac{100}{10} &&&&&& \\ \\ &&J&=&10 &&&&&& \end{array}[/latex]

This means that Joey is 10 years old and Sally is 30 years old.

For Questions 1 to 12, write and solve the equation describing the relationship.

  • The sum of two numbers is 22, and the product of these two numbers is 120. What are the numbers?
  • The difference of two numbers is 4, and the product of these two numbers is 140. What are the numbers?
  • The difference of two numbers is 8, and the sum of the squares of these two numbers are 320. What are the numbers?
  • The sum of the squares of two consecutive even integers is 244. What are these numbers?
  • The difference of the squares of two consecutive even integers is 60. What are these numbers?
  • The sum of the squares of two consecutive even integers is 452. What are these numbers?
  • Find three consecutive even integers such that the product of the first two is 38 more than the third integer.
  • Find three consecutive odd integers such that the product of the first two is 52 more than the third integer.
  • The product of the ages of Alan and Terry is 80 more than the product of their ages 4 years prior. If Alan is 4 years older than Terry, what are their current ages?
  • The product of the ages of Cally and Katy is 130 less than the product of their ages in 5 years. If Cally is 3 years older than Katy, what are their current ages?
  • The product of the ages of James and Susan in 5 years is 230 more than the product of their ages today. What are their ages if James is one year older than Susan?
  • The product of the ages (in days) of two newborn babies Simran and Jessie in two days will be 48 more than the product of their ages today. How old are the babies if Jessie is 2 days older than Simran?

Example 10.7.4

Doug went to a conference in a city 120 km away. On the way back, due to road construction, he had to drive 10 km/h slower, which resulted in the return trip taking 2 hours longer. How fast did he drive on the way to the conference?

The first equation is [latex]r(t) = 120[/latex], which means that [latex]r = \dfrac{120}{t}[/latex] or [latex]t = \dfrac{120}{r}[/latex].

For the second equation, [latex]r[/latex] is 10 km/h slower and [latex]t[/latex] is 2 hours longer. This means the second equation is [latex](r - 10)(t + 2) = 120[/latex].

We will eliminate the variable [latex]t[/latex] in the second equation by substitution:

[latex](r-10)(\dfrac{120}{r}+2)=120[/latex]

Multiply both sides by [latex]r[/latex] to eliminate the fraction, which leaves us with:

[latex](r-10)(120+2r)=120r[/latex]

Multiplying everything out gives us:

[latex]\begin{array}{rrrrrrrrr} 120r&+&2r^2&-&1200&-&20r&=&120r \\ &&2r^2&+&100r&-&1200&=&120r \\ &&&-&120r&&&&-120r \\ \hline &&2r^2&-&20r&-&1200&=&0 \end{array}[/latex]

This equation can be reduced by a common factor of 2, which leaves us with:

[latex]\begin{array}{rrl} r^2-10r-600&=&0 \\ (r-30)(r+20)&=&0 \\ r&=&30\text{ km/h or }-20\text{ km/h (reject)} \end{array}[/latex]

Example 10.7.5

Mark rows downstream for 30 km, then turns around and returns to his original location. The total trip took 8 hr. If the current flows at 2 km/h, how fast would Mark row in still water?

If we let [latex]t =[/latex] the time to row downstream, then the time to return is [latex]8\text{ h}- t[/latex].

The first equation is [latex](r + 2)t = 30[/latex]. The stream speeds up the boat, which means [latex]t = \dfrac{30}{(r + 2)}[/latex], and the second equation is [latex](r - 2)(8 - t) = 30[/latex] when the stream slows down the boat.

We will eliminate the variable [latex]t[/latex] in the second equation by substituting [latex]t=\dfrac{30}{(r+2)}[/latex]:

[latex](r-2)\left(8-\dfrac{30}{(r+2)}\right)=30[/latex]

Multiply both sides by [latex](r + 2)[/latex] to eliminate the fraction, which leaves us with:

[latex](r-2)(8(r+2)-30)=30(r+2)[/latex]

[latex]\begin{array}{rrrrrrrrrrr} (r&-&2)(8r&+&16&-&30)&=&30r&+&60 \\ &&(r&-&2)(8r&+&(-14))&=&30r&+&60 \\ 8r^2&-&14r&-&16r&+&28&=&30r&+&60 \\ &&8r^2&-&30r&+&28&=&30r&+&60 \\ &&&-&30r&-&60&&-30r&-&60 \\ \hline &&8r^2&-&60r&-&32&=&0&& \end{array}[/latex]

This equation can be reduced by a common factor of 4, which will leave us:

[latex]\begin{array}{rll} 2r^2-15r-8&=&0 \\ (2r+1)(r-8)&=&0 \\ r&=&-\dfrac{1}{2}\text{ km/h (reject) or }r=8\text{ km/h} \end{array}[/latex]

For Questions 13 to 20, write and solve the equation describing the relationship.

  • A train travelled 240 km at a certain speed. When the engine was replaced by an improved model, the speed was increased by 20 km/hr and the travel time for the trip was decreased by 1 hr. What was the rate of each engine?
  • Mr. Jones visits his grandmother, who lives 100 km away, on a regular basis. Recently, a new freeway has opened up, and although the freeway route is 120 km, he can drive 20 km/h faster on average and takes 30 minutes less time to make the trip. What is Mr. Jones’s rate on both the old route and on the freeway?
  • If a cyclist had travelled 5 km/h faster, she would have needed 1.5 hr less time to travel 150 km. Find the speed of the cyclist.
  • By going 15 km per hr faster, a transit bus would have required 1 hr less to travel 180 km. What was the average speed of this bus?
  • A cyclist rides to a cabin 72 km away up the valley and then returns in 9 hr. His speed returning is 12 km/h faster than his speed in going. Find his speed both going and returning.
  • A cyclist made a trip of 120 km and then returned in 7 hr. Returning, the rate increased 10 km/h. Find the speed of this cyclist travelling each way.
  • The distance between two bus stations is 240 km. If the speed of a bus increases by 36 km/h, the trip would take 1.5 hour less. What is the usual speed of the bus?
  • A pilot flew at a constant speed for 600 km. Returning the next day, the pilot flew against a headwind of 50 km/h to return to his starting point. If the plane was in the air for a total of 7 hours, what was the average speed of this plane?

Example 10.7.6

Find the length and width of a rectangle whose length is 5 cm longer than its width and whose area is 50 cm 2 .

First, the area of this rectangle is given by [latex]L\times W[/latex], meaning that, for this rectangle, [latex]L\times W=50[/latex], or [latex](W+5)W=50[/latex].

how to solve word problems in quadratic equation

Multiplying this out gives us:

[latex]W^2+5W=50[/latex]

[latex]W^2+5W-50=0[/latex]

Second, we factor this quadratic to get our solution:

[latex]\begin{array}{rrrrrrl} W^2&+&5W&-&50&=&0 \\ (W&-&5)(W&+&10)&=&0 \\ &&&&W&=&5, -10 \\ \end{array}[/latex]

We reject the solution [latex]W = -10[/latex].

This means that [latex]L = W + 5 = 5+5= 10[/latex].

Example 10.7.7

If the length of each side of a square is increased by 6, the area is multiplied by 16. Find the length of one side of the original square.

how to solve word problems in quadratic equation

The relationship between these two is:

[latex]\begin{array}{rrl} \text{larger area}&=&16\text{ times the smaller area} \\ (x+12)^2&=&16(x)^2 \end{array}[/latex]

Simplifying this yields:

[latex]\begin{array}{rrrrrrr} x^2&+&24x&+&144&=&16x^2 \\ -16x^2&&&&&&-16x^2 \\ \hline -15x^2&+&24x&+&144&=&0 \end{array}[/latex]

Since this is a problem that requires factoring, it is easiest to use the quadratic equation:

[latex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a},\hspace{0.25in}\text{ where }a=-15, b=24\text{ and }c=144[/latex]

Substituting these values in yields [latex]x = 4[/latex] or [latex]x=-2.4[/latex] (reject).

Example 10.7.8

Nick and Chloe want to surround their 60 by 80 cm wedding photo with matting of equal width. The resulting photo and matting is to be covered by a 1 m 2 sheet of expensive archival glass. Find the width of the matting.

how to solve word problems in quadratic equation

[latex](L+2x)(W+2x)=1\text{ m}^2[/latex]

[latex](80\text{ cm }+2x)(60\text{ cm }+2x)=10,000\text{ cm}^2[/latex]

[latex]4800+280x+4x^2=10,000[/latex]

[latex]4x^2+280x-5200=0[/latex]

Which reduces to:

[latex]x^2 + 70x - 1300 = 0[/latex]

Second, we factor this quadratic to get our solution.

It is easiest to use the quadratic equation to find our solutions.

[latex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a},\hspace{0.25in}\text{ where }a=1, b=70\text{ and }c=-1300[/latex]

Substituting the values in yields:

[latex]x=\dfrac{-70\pm \sqrt{70^2-4(1)(-1300)}}{2(1)}\hspace{0.5in}x=\dfrac{-70\pm 10\sqrt{101}}{2}[/latex]

[latex]x=-35+5\sqrt{101}\hspace{0.75in} x=-35-5\sqrt{101}\text{ (rejected)}[/latex]

For Questions 21 to 28, write and solve the equation describing the relationship.

  • Find the length and width of a rectangle whose length is 4 cm longer than its width and whose area is 60 cm 2 .
  • Find the length and width of a rectangle whose width is 10 cm shorter than its length and whose area is 200 cm 2 .
  • A large rectangular garden in a park is 120 m wide and 150 m long. A contractor is called in to add a brick walkway to surround this garden. If the area of the walkway is 2800 m 2 , how wide is the walkway?
  • A park swimming pool is 10 m wide and 25 m long. A pool cover is purchased to cover the pool, overlapping all 4 sides by the same width. If the covered area outside the pool is 74 m 2 , how wide is the overlap area?
  • In a landscape plan, a rectangular flowerbed is designed to be 4 m longer than it is wide. If 60 m 2 are needed for the plants in the bed, what should the dimensions of the rectangular bed be?
  • If the side of a square is increased by 5 units, the area is increased by 4 square units. Find the length of the sides of the original square.
  • A rectangular lot is 20 m longer than it is wide and its area is 2400 m 2 . Find the dimensions of the lot.
  • The length of a room is 8 m greater than its width. If both the length and the width are increased by 2 m, the area increases by 60 m 2 . Find the dimensions of the room.

Answer Key 10.7

Intermediate Algebra by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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how to solve word problems in quadratic equation

Solving Quadratic Equations

Many word problems Involving unknown quantities can be translated for solving quadratic equations

Methods of solving quadratic equations are discussed here in the following steps.

Step I: Denote the unknown quantities by x, y etc.

Step II: use the conditions of the problem to establish in unknown quantities.

Step III: Use the equations to establish one quadratic equation in one unknown.

Step IV: Solve this equation to obtain the value of the unknown in the set to which it belongs.

Now we will learn how to frame the equations from word problem:

1.  The product of two consecutive integers is 132. Frame an equation for the statement. What is the degree of the equation?

Method I: Using only one unknown

Let the two consecutive integers be x and x + 1

Form the equation, the product of x and x + 1 is 132.

Therefore, x(x + 1) = 132

⟹ x\(^{2}\) + x - 132 = 0, which is quadratic in x.

This is the equation of the statement, x denoting the smaller integer.

Method II: Using more than one unknown

Let the consecutive integers be x and y, x being the smaller integer.

As consecutive integers differ by 1, y - x = 1 ........................................... (i)

Again, from the question, the product of x and y is 132.

So, xy = 132 ........................................... (ii)

From (i), y = 1 + x.

Putting y = 1 + x in (ii),

x(1 + x) = 132

Solving the quadratic equation, we get the value of x. Then the value of y can be determined by substituting the value of x in y = 1 + x.

2. The length of a rectangle is greater than its breadth by 3m. If its area be 10 sq. m, find the perimeter.

Suppose, the breadth of the rectangle = x m.

Therefore, length of the rectangle = (x + 3) m.

So, area = (x + 3)x sq. m

Hence, by the condition of the problem

(x + 3)x = 10

⟹ x\(^{2}\) + 3x - 10 = 0

⟹ (x + 5)(x - 2) = 0

So, x = -5,2

But x = - 5 is not acceptable, since breadth cannot be negative.

Therefore x = 2

Hence, breadth = 2 m

and length = 5 m

Therefore, Perimeter = 2(2 + 5) m = 14 m.

x = -5 does not satisfy the conditions of the problem length or breadth can never be negative. Such a root is called an extraneous root. In solving a problem, each root of the quadratic equation is to be verified whether it satisfies the conditions of the given problem. An extraneous root is to be rejected.

Quadratic Equation

Introduction to Quadratic Equation

Formation of Quadratic Equation in One Variable

General Properties of Quadratic Equation

Methods of Solving Quadratic Equations

Roots of a Quadratic Equation

Examine the Roots of a Quadratic Equation

Problems on Quadratic Equations

Quadratic Equations by Factoring

Word Problems Using Quadratic Formula

Examples on Quadratic Equations 

Word Problems on Quadratic Equations by Factoring

Worksheet on Formation of Quadratic Equation in One Variable

Worksheet on Quadratic Formula

Worksheet on Nature of the Roots of a Quadratic Equation

Worksheet on Word Problems on Quadratic Equations by Factoring

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HOW TO SOLVE WORD PROBLEMS IN QUADRATIC EQUATIONS

We can follow the steps given below to solve word problems using quadratic equations.

Understanding the question is more important than any other thing. That is, always it is very important to understand the information given in the question rather than solving.

If it is possible, we have to split the given information. Because, when we split the given information in to parts, we can understand them easily.

Once we understand the given information clearly, solving the word problem in quadratic equation would not be a challenging work. 

When we try to solve the word problems in quadratic equations, we have to introduce "x" or some other alphabet for unknown value (=answer for our question) and form a quadratic equation with this "x". Finally we have to get value for the alphabet which was introduced for the unknown value.

If it is required, we have to draw picture for the given information. Drawing picture for the given information will give us a clear understanding about the question.

Using the alphabet introduced for unknown value, we have to translate the English statement (information) given in the question as quadratic equation equation.

In translation, we have to translate  the following English words as the corresponding mathematical symbols.

of -----> x (multiplication)

am, is, are, was, were, will be, would be --------> = (equal)

Once we have translated the English Statement (information) given in the question as quadratic equation correctly, 90% of the work will be over. The remaining 10% is just getting the answer. That is solving for the unknown. 

These are the steps most commonly involved in solving word problems in quadratic equations.

Let us see how the above explained steps work in solving word problems using quadratic equations. 

A piece of iron rod cost $ 60. If the rod was 2 meter shorter and each meter costs $ 1 more and the total cost  would remain unchanged. What is the length of the rod?  

Let us understand the given information. There are three information given in the question. 

1.  A piece of iron rod costs $ 60.

2.  If the rod was 2 meter shorter and each meter costs $ 1 more

3.  Total cost  would remain unchanged.

Target of the question : What is the length of the rod?

Step 3 :

Let "x" be the length of the rod.

Clearly, we have to find the value of "x" 

If the rod is 2 meter shorter, length of the rod is

=  (x-2)

From the third information, we have the following statements.

Total cost of rod having length x meters is $ 60.

Total cost of rod having length (x-2) meters is $ 60.

Cost of 1 meter of rod having length x meters is

=  60 / x  -----(1)

Cost of 1 meter of rod having length (x-2) meters is

=  60 / (x - 2) -----(2)

From the second information, we can consider the following example. 

That is, if the cost of 1 meter of rod x is $10, then the cost of 1 meter of rod (x-2) will be $11.  

$10 &  $11 can be balanced as shown below.

10 + 1  =  11

(This is just for en example)

If we apply the same logic for (1) & (2), we get 

(60 / x) + 1  =  60 / (x - 2)

(60 + x) / x  =  60 / (x - 2)

(x + 60)(x - 2)  =  60x

x 2  + 58x - 120  =  60x

x 2  - 2x - 120  =  0

(x - 12)(x + 10)  =  0 

x - 12  =  0     or     x + 10  =  0

x  =  12     or     x  =  -10

Because length can never be a negative value, we can ignore x  =  -10. 

x  =  12

So, the length of the rod is 12 meter. 

Would you like to practice more word problems in quadratic equations ? 

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Kindly mail your feedback to   [email protected]

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Need Help Solving Those Dreaded Word Problems Involving Quadratic Equations?

Yes, I know it's tough. You've finally mastered factoring and using the quadratic formula and now you are asked to solve more problems!

Except these are even more tough. Now you have to figure out what the problem even means before trying to solve it. I completely understand and here's where I am going to try to help!

There are many types of problems that can easily be solved using your knowledge of quadratic equations. You may come across problems that deal with money and predicted incomes (financial) or problems that deal with physics such as projectiles. You may also come across construction type problems that deal with area or geometry problems that deal with right triangles.

Lucky for you, you can solve the quadratic equations, now you just have to learn how to apply this useful skill.

On this particular page, we are going to take a look at a physics "projectile problem".

Projectiles - Example 1

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -16t 2 + 40ft + 1.5. Find the maximum height attained by the ball.

Let's first take a minute to understand this problem and what it means. We know that a ball is being shot from a cannon. So, in your mind, imagine a cannon firing a ball. We know that the ball is going to shoot from the cannon, go into the air, and then fall to the ground.

So, here's a mathematical picture that I see in my head.

Now let's talk about what each part of this problem means. In our equation that we are given we must be given the value for the force of gravity (coefficient of t 2 ). We must also use our upward velocity (coefficient of t) and our original height of the cannon/ball (the constant or 1.5). Take a look...

Now that you have a mental picture of what's happening and you understand the formula given, we can go ahead and solve the problem.

  • First, ask yourself, "What am I solving for?" "What do I need to find?" You are asked to find the maximum height (go back and take a look at the diagram). What part of the parabola is this? Yes, it's the vertex! We will need to use the vertex formula and I will need to know the y coordinate of the vertex because it's asking for the height.
  • Next Step: Solve! Now that I know that I need to use the vertex formula, I can get to work.

Just as simple as that, this problem is solved.

Let's not stop here. Let's take this same problem and put a twist on it. There are many other things that we could find out about this ball!

Projectiles - Example 2

Same problem - different question. Take a look...

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -16t 2 + 40ft + 1.5. How long did it take for the ball to reach the ground?

Now, we've changed the question and we want to know how long did it take the ball to reach the ground.

What ground, you may ask. The problem didn't mention anything about a ground. Let's take a look at the picture "in our mind" again.

Do you see where the ball must fall to the ground. The x-axis is our "ground" in this problem. What do we know about points on the x-axis when we are dealing with quadratic equations and parabolas?

Yes, the points on the x-axis are our "zeros" or x-intercepts. This means that we must solve the quadratic equation in order to find the x-intercept.

Let's do it! Let's solve this equation. I'm thinking that this may not be a factorable equation. Do you agree? So, what's our solution?

Hopefully, you agree that we can use the quadratic formula to solve this equation.

The first time doesn't make sense because it's negative. This is the calculation for when the ball was on the ground initially before it was shot.

This actually never really occurred because the ball was shot from the cannon and was never shot from the ground. Therefore, we will disregard this answer.

The other answer was 2.54 seconds which is when the ball reached the ground (x-axis) after it was shot. Therefore, this is the only correct answer to this problem.

Ok, one more spin on this problem. What would you do in this case?

Projectiles - Example 3

A ball is shot from a cannon into the air with an upward velocity of 40 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -16t 2 + 40ft + 1.5. How long does it take the ball to reach a height of 20 feet?

Yes, this problem is a little trickier because the question is not asking for the maximum height (vertex) or the time it takes to reach the ground (zeros), instead it it asking for the time it takes to reach a height of 20 feet.

Since the ball reaches a maximum height of 26.5 ft, we know that it will reach a height of 20 feet on the way up and on the way down.

Let's just estimate on our graph and also make sure that we get this visual in our head.

From looking at this graph, I would estimate the times to be about 0.7 sec and 1.9 sec. Do you see how the ball will reach 20 feet on the way up and on the way down?

Now, let's find the actual values. Where will we substitute 20 feet?

Yes, we must substitute 20 feet for h(t) because this is the given height. We will now be solving for t using the quadratic formula. Take a look.

Our actual times were pretty close to our estimates. Just don't forget that when you solve a quadratic equation, you must have the equation set equal to 0. Therefore, we had to subtract 20 from both sides in order to have the equation set to 0.

You've now seen it all when it comes to projectiles!

Great Job! Hopefully you've been able to understand how to solve problems involving quadratic equations. I also hope that you better understand these common velocity equations and how to think about what this problem looks like graphically in order to help you to understand which process or formula to use in order to solve the problem.

  • Quadratic Equations
  • Projectile Problems

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