A ball thrown upwards from a rooftop, 80ft above the ground. It will reach a maximum vertical height and then fall back to the ground. The

Question

 

height of the ball from the ground at time t is h, which is given by, h=-16t²+64t+80

 

a. What is the height reached by the ball after 1 second?

b. What is the maximum height reached by the ball?

c. How long will it take before hitting the ground?

Answer 1

a. We use the equation provided, plugging in t=1. We get h=-16+64+80=128.

 

b. Whenever you see the word "maximum" (or "minimum," in a different setting) think "set derivative equal to zero/undefined." To determine the maximum height, we need to derivative of the height equation:

 

h' = -32t + 64

 

Set h'=0 and solve for t:

 

0=-32t+64

t=2

 

Thus, maximum height occurs at t=2. Plug this in to the original equation for h:

 

h=-16*(2)^2+64*2+80=144

 

(This answer looks good because it's bigger than our answer for part a! Having a basic understanding of what the ball is doing can save you from making silly mistakes.)

 

c. The ground is height h=0. So, we solve for t:

 

0=-16t^2+64t+80

 

We use the quadratic formula to find t=-1 and t=5. But, t=-1 doesn't make any sense (-1 second?) so our solution is t=5.

 

I'm available for online tutoring! :)

Answer 2

a. -16(1)² + 64(1) + 80

     -16 + 64 + 80 = 128

 

b. x = -b/2a = -64/(2(-16)) = 2

y = -16(2)² + 64(2) + 80

      -64 + 128 + 80 = 144

 

c. -16t² + 64t + 80 = 0

    t² - 4t - 5 = 0

    (t - 5)(t + 1) = 0

 

t - 5 = 0

t = 5

 

t + 1 = 0

t = -1. Can't be an answer.