Answer to Question #236132 in Physics for Nathan

Question

A ball is thrown straight downward with an initial velocity of 0.5 m/s down from a height of 4 meters. What is the magnitude and direction of the ball's velocity 0.7 seconds after it is released?

Answer 1

We can find the ball's velocity 0.7 seconds after it is released from the kinematic equation:

$$v=v_0-gt,$$ $$v(t=0.7\ s)=-0.5\ \dfrac{m}{s}-9.8\ \dfrac{m}{s^2}\cdot0.7\ s=-7.36\ \dfrac{m}{s}.$$

The sign minus means that the ball's velocity directed downward.

Answer:

$v(t=0.7\ s)=7.36\ \dfrac{m}{s}$, downward.