Please help me out

Question

A rock is thrown upward with a speed of 48 feet per second from the edge of a cliff 400 feet above the ground. What is the speed of the rock when it hits the ground? Use acceleration due to gravity as -32 feet per second squared and approximate your answer to 3 decimal places.

Answer 1

v(t) = v₀ + at

v₀ = 48 ft/s

a = -32 ft/s²

v(t) = 48 - 32t

You can find how long it takes to reach the ground, and then plug that time into the equation above. It reaches the ground when h(t)=0.

h(t) = -16t² + 48t + 400 = 0

Solve for t using the quadratic formula. You will get a positive and a negative value. The positive value is the correct time. Plug the positive value of t into

v(t) = 48 - 32t,

and that will give the velocity at the bottom. It will be a negative number, which means it is going down, as expected.