Answer:
a) ![t = \sqrt{\frac{h}{2g}}](https://tex.z-dn.net/?f=t%20%3D%20%5Csqrt%7B%5Cfrac%7Bh%7D%7B2g%7D%7D)
b) Ball 1 has a greater speed than ball 2 when they are passing.
c) The height is the same for both balls = 3h/4.
Explanation:
a) We can find the time when the two balls meet by equating the distances as follows:
Where:
: is the initial height = h
: is the initial speed of ball 1 = 0 (it is dropped from rest)
(1)
Now, for ball 2 we have:
Where:
: is the initial height of ball 2 = 0
(2)
By equating equation (1) and (2) we have:
![h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}](https://tex.z-dn.net/?f=%20h%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E%7B2%7D%20%3D%20v_%7B0_%7B2%7D%7Dt%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E%7B2%7D%20)
![t=\frac{h}{v_{0_{2}}}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bh%7D%7Bv_%7B0_%7B2%7D%7D%7D)
Where the initial velocity of the ball 2 is:
![v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh](https://tex.z-dn.net/?f=v_%7Bf_%7B2%7D%7D%5E%7B2%7D%20%3D%20v_%7B0_%7B2%7D%7D%5E%7B2%7D%20-%202gh)
Since
= 0 at the maximum height (h):
![v_{0_{2}} = \sqrt{2gh}](https://tex.z-dn.net/?f=v_%7B0_%7B2%7D%7D%20%3D%20%5Csqrt%7B2gh%7D)
Hence, the time when they pass each other is:
![t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}](https://tex.z-dn.net/?f=%20t%20%3D%20%5Cfrac%7Bh%7D%7B%5Csqrt%7B2gh%7D%7D%20%3D%20%5Csqrt%7B%5Cfrac%7Bh%7D%7B2g%7D%7D%20)
b) When they are passing the speed of each one is:
For ball 1:
![v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}](https://tex.z-dn.net/?f=%20v_%7Bf_%7B1%7D%7D%20%3D%20-%20gt%20%3D%20-g%2A%5Csqrt%7B%5Cfrac%7Bh%7D%7B2g%7D%7D%20%3D%20-%200.71%5Csqrt%7Bgh%7D%20)
The minus sign is because ball 1 is going down.
For ball 2:
![v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}](https://tex.z-dn.net/?f=v_%7Bf_%7B2%7D%7D%20%3D%20v_%7B0_%7B2%7D%7D%20-%20gt%20%3D%20%5Csqrt%7B2gh%7D%20-%20g%2A%5Csqrt%7B%5Cfrac%7Bh%7D%7B2g%7D%7D%20%3D%20%28%5Csqrt%7B1%7D%20-%20%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%29%2A%5Csqrt%7Bgh%7D%20%3D%200.41%5Csqrt%7Bgh%7D)
Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.
c) The height of the ball is:
For ball 1:
![y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h](https://tex.z-dn.net/?f=%20y_%7B1%7D%20%3D%20h%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E%7B2%7D%20%3D%20h%20-%20%5Cfrac%7B1%7D%7B2%7Dg%28%5Csqrt%7B%5Cfrac%7Bh%7D%7B2g%7D%7D%29%5E%7B2%7D%20%3D%20%5Cfrac%7B3%7D%7B4%7Dh%20)
For ball 2:
![y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h](https://tex.z-dn.net/?f=%20y_%7B2%7D%20%3D%20v_%7B0_%7B2%7D%7Dt%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E%7B2%7D%20%3D%20%5Csqrt%7B2gh%7D%2A%5Csqrt%7B%5Cfrac%7Bh%7D%7B2g%7D%7D%20-%20%5Cfrac%7B1%7D%7B2%7Dg%28%5Csqrt%7B%5Cfrac%7Bh%7D%7B2g%7D%7D%29%5E%7B2%7D%20%3D%20%5Cfrac%7B3%7D%7B4%7Dh%20)
Then, when they are passing the height is the same for both = 3h/4.
I hope it helps you!