Question

A ball is thrown upward with a speed of 40 feet per second from the edge of a cliff 500 feet above the ground. What is the speed of the ball 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

use one of the equations of constant acceleration
initial velocity u = 40 , a = -32, v = final velocy and displacement s = -500

v^2 = u^2 + 2as

find v

Answer 2

plugging in the values

v^2 = 4^2 + 2 * -32 * -500

Answer 3

sorry that should be 40^2
v^2 = 40^2 + 2 * -32 * -500