Question

A stone is launched vertically upward from a cliff 192 ft above the ground at a speed of 16 ft/s. Its height above the ground t seconds after the launch is given by s = -16t² + 16t + 192 for 0≤t≤4. When does the stone reach its maximum​ height?

Answer 1

Answer:

0.5 sec

Step-by-step explanation:

s(t) = -16t² + 16t + 192

Lets differentiate the equation first w.r.t 't', we will have:

s'(t)=-32t + 16

In order to find critical points, equating s'(t) to zero, and determine values where s'(t) is undefined

Therefore, s'(t)=0

-32t + 16= 0     (solving for t)

-32t = -16

t= 16/32

t= 0.5s

Now, we will evaluate 's' at endpoints and critical points.

->s(0)= -16(0)² + 16(0) + 192

s(0)= 192

->s(0.5)= -16(0.5)² + 16(0.5) + 192

s(0.5)= 196

->s(4)= -16(4)² + 16(4) + 192

s(4)= 0

Thus, s(0.5)= 196 is the largest function from above i.e the absolute max. value of s on [0,4]

the stone reach its maximum​ height i.e 196 ft at 0.5sec