Do the Balls Cross Paths at 1/2h, Above 1/2h, or Below 1/2h?

[kbaumen]

Homework Statement

A ball is thrown straight upward with an initial speed v0. When it reaches
the top of its flight at height h, a second ball is thrown straight upward with
the same initial speed. Do the balls cross paths (a) at height 1⁄2h (b) above
1⁄2h or (c) below 1⁄2h. Explain your reasoning.

The Attempt at a Solution

My approach was quite simplistic. Since the second ball has an initial speed, over the given time interval, it'll move over a longer distance than the first one. That was also what the answers said. However, I don't find this answer satisfying. I would like to prove this analytically but unfortunately I can't find a way to do so.

I tried to write out the distance equations for both balls:

$$h=v_0t - \frac{gt^2}{2}$$
$$h=h_t - \frac{gt^2}{2}$$

where $$h$$ is the height at which the balls cross paths, $$h_t$$ is the max height the first ball reaches and $$v_0$$ is the initial speed. Now the RHS's of both equations are equal, so equating them and solving for t gives

$$t=\frac{h_t}{v_0}$$

Substituting that back into the first of the two equations gives

$$h=h_t-\frac{gh_t^2}{2v_0^2}$$

From here, I can't get any further. I'm not even sure this is the right approach.

Thank you for any help.

 

[Fightfish]

Right, I'm a bit too lazy to mull over your working, so I might be kinda starting off almost anew.

First thing, the maximum height that the ball reaches can be easily represented in terms of other variables. So,

$$h_{max} = \frac{v_{0}^{2}}{2g}$$​

Now, let the balls meet after a time t. Then, the distances traveled by ball 1 and ball 2 respectively are

$$d_{1} = \frac{1}{2}gt^{2}$$
$$d_{2} = v_{0}t - \frac{1}{2}gt^{2}$$
​

We further know that the total distance must be equal to the max height attained by ball 1.

$$d_{1} + d_{2} = v_{0}t = h_{max} = \frac{v_{0}^{2}}{2g}$$​

Then,

$$t = \frac{v_{0}}{2g}$$​

Which gives

$$d_{2} - d_{1} = v_{0}t - gt^{2} = \frac{v_{0}^{2}}{2g} - \frac{v_{0}^{2}}{4g} = \frac{v_{0}^{2}}{4g} > 0$$
​

Hence, we are done in showing that d2 is greater than d1.

 

[kbaumen]

Oh, that makes sense.

Cheers mate.