Question

A lead ball is dropped into a lake from a diving board 5.0 m above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 3.0 s after it is released. How deep is the lake

Answer 1

Answer:

$|D_{depth} |=19.697m$

Explanation:

To find Depth D of lake we must need to find the time taken to hit the water.So we use equation of simple motion as:

Δx=vit+(1/2)at²

$x_{f}-x_{i}=v_{i}t+(1/2)at^{2}\\ -5.0m=(o)t+(1/2)(-9.8m/s^{2} )t^{2}\\ -4.9t^{2}=-5.0\\ t^{2}=5/4.9\\t=\sqrt{1.02} \\t=1.01s$

As we have find the time taken now we need to find the final velocity vf from below equation as

$v_{f}=v_{i}+at\\v_{f}=0+(-9.8m/s^{2} )(1.01s) \\v_{f}=-9.898m/s$

So the depth of lake is given by:

first we need to find total time as

t=3.0-1.01 =1.99 s

$|D_{depth} |=|vt|\\|D_{depth} |=|(-9.898m/s)(1.99s)|\\|D_{depth} |=19.697m$