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+80h=−16t2+64t+80. Find the time it will take before hitting the ground. Consider the following HINTS:
1. When the ball hits the ground, the height is zero.
2. Use the concept of solving quadratic equations to solve this problem.
3. The time is "positive".
Answers
Step-by-step explanation:
. 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.