A stone is thrown straight up

Question

A stone is thrown straight up from the edge of a roof, 600 feet above the ground, at a speed of 16 feet per second.

A. Remembering that the acceleration due to gravity is -32 feet per second squared,

how high is the stone 3 seconds later?   

b. What time does the stone hit the ground?

c. What is the velocity of the stone when it hits the ground.

Answer 1

a(t) = -32 = acceleration at time t

v(t) = velocity at time t = ∫(-32)dt = -32t + C

Since v(0) = 16, -32(0) + C = 16. So, C = 16. Therefore, v(t) = -32t + 16.

h(t) = height at time t = ∫v(t)dt = ∫(-32t + 16)dt = -16t² + 16t + C₁

Since h(0) = 600, C₁ = 600. So, h(t) = -16t² + 16t + 600.

Height after 3 seconds = h(3).

Stone hits the ground when h(t) = 0.

To find velocity when the stone hits the ground , evaluate v(t) at the value of t obtained from solving the equation h(t) = 0.