- #1

opus

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## Homework Statement

You throw a ball straight upward with initial speed ##v_0##. How long does it take to return to your hand?

## Homework Equations

##v=v_0+at##

##x-x_0=v_0t+\frac{1}{2}at^2##

## The Attempt at a Solution

My question is a general one that relates to the described solution.

According to the text, the time to reach the top of the flight is ##t=\frac{v_0}{g}##.

It then goes on to say that the downward motion takes the same time as the upward motion, so the total time back to the ground is ##\frac{2v_0}{g}##.

But why would the time to the ground be the same as the time to the top? My confusion lies in the fact that on the way up, we have an initial velocity not equal to zero. On the way down, the initial velocity

*is*zero, so it wouldn't start out going as fast as the first part of the parabola. Then I thought that the first part of the parabola, although it has an initial velocity not equal to zero, does slow down throughout it's entire flight up. In contrast, on the way down, it has an initial velocity of zero but is speeding up until it hits the ground.

So my question is, do these two halves of the parabola always balance out? That is, will the initial velocity of the first half of the flight, be equal to the final velocity of the second half of the flight?