Question

A bowling ball rolls off the edge of a flat roof at a velocity of 16 m/s. What is the actual speed of the ball one second later?

Answer 1

Answer:

The horizontal velocity is constant at 16 m/s.

After 1 sec since v = a t then 9.8 m/s^2 * 1 sec = 9.8 m/s for the vertical velocity

V = (Vx^2 + Vy^2)^1/2 = (16^2 + 9.8^2)^1/2 = 18.8 m/s