height of the ball

Question

If a baseball is projected upward from ground level with an initial velocity of
128
feet per? second, then its height is a function of ?time, given by
s=−16tsquared+ 128t.


What is the maximum height reached by the? ball?

Answer 1

Hey Shawn,

We are given V initial = 128 feet/s

s=-16t²+128t

what we know is that at maximum height v=0

we also know that v=ds/dt

we differentiate the equation s=-16t^2+128t with respect to t (time) to obtain Velocity equation

v=-32t+128

set v=0 (at max height)

so we the equation -32t+128=0 <---- solve for t 

t=-128/-32

t=4s

go back to the first equation 

s=-16t^2+128t replace t by 4

s=-16(4)^2+128(4)

s=-256+512

s=256 feet