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A stone is launched vertically upward from a cliff 384 ft above the ground at a speed of 80 ft divided by s. Its height above the ground t seconds after the launch is given by s equals negative 16 t squared plus 80 t plus 384 for 0 less than or equals t less than or equals 8. When does the stone reach its maximum​ height?

Respuesta :

estryzy

Answer:

2.5 seconds

Explanation:

s(t) = -16t^2 + 80t + 384

for

0≤t≤8

First we differentiate s(t) to get s'(t)

s'(t) = -32t + 80

Let us then find the critical point; thus we will equate s'(t) to zero and then search for values where s'(t) is undefined

s'(t) = -32t + 80 = 0

t = 80/32

t = 2.5 sec

Let us evaluate s at the critical points and end points

s(0) = -16(0)^2 + 80(0) + 384 = 384

s(2.5) = -16(2.5)^2 + 80(2.5) + 384 = 684

s(8) = -16(8)^2 + 80(8) + 384 = 0

Thus, the stone attains it maximum height of 684ft at at t=2.5s

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