Question

If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t − 16t^2.

What is the maximum height reached by the ball?

What is the velocity of the ball when it is 384 ft above the ground on its way up?

What is the velocity of the ball when it is 384 ft above the ground on its way down?

Answer 1

Answer:

400ft.    32ft/s       -32ft/s

Explanation:

In reality the gravitational acceleration is 9.81 so the quadratic coefficient of the function should be 9.81/2

Anyway for the sake of assumtion let us takes=160t-16t^2

 

ds/dt=160-32t=0

 

t=160/32= 5 seconds.

s=160*160/32-16*(160/32)^2= 400 mts

 

 

s=384 mts

160t-16t^2=384

i.e

16t^2-160t+384=0

 

t^2-10t+24=0

(t-6)(t-4)=0

t=[4,6]

we have to take t=4 because it is all the up i.e <5

 

velocity =v=ds/dt=160-32t

 

v=160-32*4=32 ft/sec still going up

 

for all the way down take t=6 whuch is >5

 

v=160-6*32=-32 ft/sec (falling down!!!)