Answer:
a) [tex]t = \sqrt{\frac{h}{2g}}[/tex]
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:
[tex] y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2} [/tex]
Where:
[tex]y_{0_{1}}[/tex]: is the initial height = h
[tex] v_{0_{1}}[/tex]: is the initial speed of ball 1 = 0 (it is dropped from rest)
[tex] y = h - \frac{1}{2}gt^{2} [/tex] (1)
Now, for ball 2 we have:
[tex]y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}[/tex]
Where:
[tex]y_{0_{2}}[/tex]: is the initial height of ball 2 = 0
[tex]y = v_{0_{2}}t - \frac{1}{2}gt^{2}[/tex] (2)
By equating equation (1) and (2) we have:
[tex] h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} [/tex]
[tex]t=\frac{h}{v_{0_{2}}}[/tex]
Where the initial velocity of the ball 2 is:
[tex]v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh[/tex]
Since [tex]v_{f_{2}}^{2}[/tex] = 0 at the maximum height (h):
[tex]v_{0_{2}} = \sqrt{2gh}[/tex]
Hence, the time when they pass each other is:
[tex] t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}} [/tex]
b) When they are passing the speed of each one is:
For ball 1:
[tex] v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh} [/tex]
The minus sign is because ball 1 is going down.
For ball 2:
[tex]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}[/tex]
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:
[tex] y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h [/tex]
For ball 2:
[tex] 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 [/tex]
Then, when they are passing the height is the same for both = 3h/4.
I hope it helps you!
