Question

A lead ball is dropped into a lake from a diving board 5m above the water . It hits the water with a certain velocity and then sink with constant velocity in bottom . It reaches the bottom 5 second after it is dropped If g= 10m/s2 . Find depth of the lake and average velocity of the ball ?

Answer 1

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:

$U=mgh$

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:

$K=\frac{1}{2}mv^2$

where m is the mass and v is the speed. Equalizing the two terms, we have:

$mgh=\frac{1}{2}mv^2\\2gh=v^2\\v=\sqrt{2gh}=\sqrt{2(10 m/s^2)(5 m)}=10 m/s$

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

$d=vt=(10 m/s)(5 s)=50 m$

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:

$v=\frac{d}{t}$

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

$S=\frac{1}{2}gt_a^2\\t_a = \sqrt{\frac{2S}{g}}=\sqrt{\frac{2(5 m)}{10 m/s^2}}=1 s$

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

$v=\frac{55 m}{6 s}=9.2 m/s$