Question

If a ball is thrown vertically upward with a velocity of 96 ft/s, then its height after t seconds is s = 96t - 16t2.
(a) What is the maximum height reached by the ball?
1) ft

(b) What is the velocity of the ball when it is 96 ft above the ground on its way up? (Consider up to be the positive direction. Round the answer to one decimalplace.)
2) ft/s
What is the velocity of the ball when it is 96 ft above the ground on its way down? (Round the answer to one decimal place.)
3) ft/s

Answer 1

If a ball is thrown vertically upward with a velocity of 96 ft/s, then its height after t seconds is s = 96t - 16t2.

vi=96ft/s

s=96t-16t2

then taking derivative to find velocity:

v=96-32t

(a) What is the maximum height reached by the ball?
the maximum height reached happens when the ball stops going up and starts going down, in other words, velocity is zero

0=96-32t >>>32t=96 >>>t=3

then s= 96t-16t2=288-144=144ft

(b) What is the velocity of the ball when it is 96 ft above the ground on its way up? (Consider up to be the positive direction. Roundthe answer to one decimal place.)
you need to find the time when the ball reaches 96ft

s=96t-16t2>>>96=96t-16t2

then set it equal to zero and solve:0=t2-6t+6

t=4.7321, 1.2679 and it has to be on the way up, so we take the lesser value and use the velocity formula:

v=96-32t=96-40.5728=55.4272ft/s=55.4ft/s

What is the velocity of the ball when it is 96 ft above the ground on its way down? (Round the answer to one decimalplace.)
3) ft/s

now we take the second time from question 2

v=96-32t=96-151.4272=-55.4ft/s or 55.4ft/s going down