Question

A rock is thrown vertically upward from the ground, with an initial velocity of 64 ft/sec. Its position function is s(t) = –16t2 + 64t. What is its velocity in ft/sec when it hits the ground?

Answer 1

Velocity of rock when it hits the ground is -64 ft/s

Step-by-step explanation:

Its position function is s(t) = –16t² + 64t

When it reaches back ground s(t) = 0 ft

Substituting

s(t) = –16t² + 64t = 0

t² - 4t = 0

t² = 4t

t =4 seconds

Now we need to find velocity when it reaches ground, that is velocity after 4 seconds.

Differentiating s(t) equation

v(t) = -32t+64

Substituting t = 4 seconds

v(4) = -32 x 4+64 = -64 ft/s

Velocity of rock when it hits the ground is -64 ft/s