Answer to Question #236132 in Physics for Nathan

Question #236132

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?


1
Expert's answer
2021-09-12T19:00:10-0400

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


v=v0gt,v=v_0-gt,v(t=0.7 s)=0.5 ms9.8 ms20.7 s=7.36 ms.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 msv(t=0.7\ s)=7.36\ \dfrac{m}{s}, downward.


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