Mike Trout is standing on the second deck of a baseball stadium that is 80 feet tall. He hits a baseball to analyze its parabolic path.

Question

Mike Trout is standing on the second deck of a baseball stadium that is 80 feet tall. He hits a baseball to analyze its parabolic path. The equation that best represents the path of the ball is         , where  is the time, in seconds, and  is the height of the ball above ground. For what interval is the ball higher than 128 feet in the air?

A. -1 < t < 5

B. t < -1 and t > 5

C.t < 1 and t > 3

  

  

  

D. 1 < t < 3

For what interval is the ball higher than 128 ft in the air? s the height of the ball above ground. For what interval is the ball higher than 128 ft in the air?s the height of the ball above ground. For what interval is the ball higher than 128 ft in the air?s the height of the ball above ground. For what interval is the ball higher than 128 ft in the air? s the height of the ball above ground. For what interval is the ball higher than 128 ft in the air?   is the height of the ball above ground. For what interval is the ball higher than 128 ft in the air? 

Answer 1

if s(t) = -16t² + 64t + 80

then s(t) = -16(t² - 4t - 5) = -16(t - 5) (t + 1)

=> t - 5 = 0 and t + 1 = 0

=> t = 5 seconds and t = -1 seconds when height = 0

0 to 0 is 6 seconds, so max height at -1 + 3 = 2 seconds. max height = -16 (-3)(3) = 144 feet

this tells us that the baseball is over 128 feet is between 1 and 5 seconds (actually 2 +/- 1 approximately), and is continuous period (goes up and comes down), so we can eliminate choices (A) interval 0 to 0 height, (B) below 0 height and (C) discontinuous ranges (upside down parabola). Hence answer must be D.

So we don't need to solve the equation for height = 128, but we can simplify to double check the answer.

-16(t - 5) (t + 1) = 128

=> (t - 5) (t + 1) = -8

for t =1 (-4)(2) = -8 and for t = 3 (-2)(4) = -8

Answer 2

Substitute 128 ft in the given equation and solve for t. You get a quadratic equation which looks like -16t² +64t -48 = 0 This can be solved using the quadratic formula which results in 2 roots--1 second and 3 seconds. In less than 1 second the ball hasn't reached 128 ft and greater than 3 seconds, the ball has fallen below 128 ft. The correct answer for this is D.

Answer 3

You can start by eliminating choices a and b since a -t value would reunite you to go back in time to before the ball was thrown

To see when the ball is higher than 128 feet solve

-16t^2 +64t +80 =128 ( the equation you listed in your question should be -16t^2 +64t +80 (not 80t)

-16t^2 +64t -48=0 Multiply both sides by -1

16t^2 -64t+48=0 divide both sides by 16

t^2 -4t +4=0

(t-3)(t-1)= 0 t=1 t=3 so the ball is higher than 128 feet between 1 and 3 seconds ( This is a downward opening parabola)

That would be choice D

Answer 4

T has to be less than 5 because gravity takes over.

T is over -1 because the ball is being pitched.

Your answer is either a or d.

Depends on the pressure/speed of the hit.

Answer 5

Let's start by realizing the equation -16t² + 64t + 80t is incorrect.

The equation should read -16t + 64t + 80.

There are 2 ways you can work this problem. Decide which is easier for you and just go with it. I will walk you through both.

First, looking at your answer choice, it is obvious that A and B are incorrect. Time can never be negative. Therefore you are only left with C or D.

The easiest way to answer this question is to find the vertex of the parabola.

You remember from Algebra 1.

The vertex (x,y) is found with the following:

X = -b/2a. a=-16, b=64, c=80

X = -64 / 2(-16) = -64 / -32 = 2

Substitute 2 for t in your equation = -16t² + 64t +80 =

-16(4) + 64(2) + 80 =

-64 + 128 + 80 = 144

144 is the y in your vertex. So the vertex is (2, 144).

This means that in 2 seconds the height of the ball will be at 144 feet (the highest height it will reach.

So, since 144 > 128, 2 seconds must be within the interval of time.

Between C and D, only D allows for 2 to be in your answer.

Therefore, 1< t < 3.

You can also workout the equation -16t² + 64t + 80 = 128

This will tell you when the ball will be at 128 feet.

Subtract 128 from both sides and your equation becomes

-16t² +64t -48 = 0

Factor out a -16 and now your equation is

t² - 4t + 3 = 0

(Remember to change your signs when you divide by a negative)

Factoring the equation we get (t - 1)(t - 3) = 0

Therefore t = 1 or t = 3.

This tells us that at 1 second and at 3 seconds the ball will be AT 128 feet.

At 1 second, the ball rises to 128 feet. At 3 seconds, the ball is on its way back down, and will reach 128 feet once again.

Therefore, the ball is higher than 128 feet between 1 and 3 seconds.

1 < t < 3.

Hope this helps.