1. Depth of the lake: 50 m
Explanation:
The first part of the problem can be solved by using conservation of energy.
When it is dropped, all the mechanical energy of the ball is potential energy, given by:
![U=mgh](https://tex.z-dn.net/?f=U%3Dmgh)
where m is the mass, g is the gravitational acceleration and h is the height.
When the bal hits the water, all the mechanical energy has been converted into kinetic energy:
![K=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
where m is the mass and v is the speed. Equalizing the two terms, we have:
![mgh=\frac{1}{2}mv^2\\2gh=v^2\\v=\sqrt{2gh}=\sqrt{2(10 m/s^2)(5 m)}=10 m/s](https://tex.z-dn.net/?f=mgh%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2%5C%5C2gh%3Dv%5E2%5C%5Cv%3D%5Csqrt%7B2gh%7D%3D%5Csqrt%7B2%2810%20m%2Fs%5E2%29%285%20m%29%7D%3D10%20m%2Fs)
Then the ball travels in the water, keeping this constant velocity, for t=5 s. So, the total distance traveled underwater (which is the depth of the lake) is
![d=vt=(10 m/s)(5 s)=50 m](https://tex.z-dn.net/?f=d%3Dvt%3D%2810%20m%2Fs%29%285%20s%29%3D50%20m)
2. Average velocity: 9.2 m/s
Explanation:
The average velocity during the whole path of the ball is equal to the ratio between the total distance traveled and the total time taken:
![v=\frac{d}{t}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bd%7D%7Bt%7D)
We already know the total distance: 5 meters above the water and 50 m underwater, so
d = 5 m + 50 m = 55 m
For the total time, we need to calculate the time spent above the water, which is given by
![S=\frac{1}{2}gt_a^2\\t_a = \sqrt{\frac{2S}{g}}=\sqrt{\frac{2(5 m)}{10 m/s^2}}=1 s](https://tex.z-dn.net/?f=S%3D%5Cfrac%7B1%7D%7B2%7Dgt_a%5E2%5C%5Ct_a%20%3D%20%5Csqrt%7B%5Cfrac%7B2S%7D%7Bg%7D%7D%3D%5Csqrt%7B%5Cfrac%7B2%285%20m%29%7D%7B10%20m%2Fs%5E2%7D%7D%3D1%20s)
So the total time is: 1 second above the water + 5 seconds underwater:
t = 1 s + 5 s = 6 s
Therefore, the average velocity is
![v=\frac{55 m}{6 s}=9.2 m/s](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B55%20m%7D%7B6%20s%7D%3D9.2%20m%2Fs)