1. A ball is thrown upwards from a rooftop, 80 m above the ground. It will reach a maximum vertical height and then fall back to the ground. The height of the ball from the ground at time t is h, and is given by h = -16t2 + 64t + 80. Find:
a. The height reached by the ball after 1 second?
b. The maximum height reached by the ball?
c. The time it will take before hitting the ground?
Answers
Answer:
Step-by-step explanation:
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.