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?

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

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