A ball is shot directly upward, and then it comes back to the place where it was shot. Suppose we have air resistance. Suppose $t_1$ is the time period from the moment that the ball was shot to the moment that it reached its highest altitude, and $t_2$ is the time period from the moment it reached its highest altitude to the moment it reached its original position. Is $t_1=t_2$? Why or why not?
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2 Answers
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Let us consider the velocities of the ball at some interim height $h$ on the ball's way up and down. The absolute value of the velocity is higher on the ball's way up, as the potential energy of the ball is the same and the total energy is less on its way down due to air resistance. As this applies to any height $h$, the conclusion is that the time the ball goes up $t_1$ is less than the time the ball goes down $t_2$.
answered Oct 14, 2012 at 18:22
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In addition to the gravitational force, we now have the drag force. On the way up, both these forces act downward. So the ball experiences an increased net acceleration. This implies that the instantaneous velocity will take less time to decrease to zero at maximum height. On the way back, both the forces act in opposite directions so the net acceleration is reduced. It will now take longer for the instantaneous velocity to increase to its initial velocity (which is to conserve the energy.) Thus, t2>t1.
