Answer
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Hint: As a first step, you could find the time taken by the ball to hit the surface. Then you could go for finding the velocity with which the ball hits the surface of the lake. After that you could assume the ball will have the same velocity while falling through the depth of the lake and thus find the depth.
Complete Step by step solution:
In the question, we are told that a lead ball is dropped into the lake from a height of 5m from a diving board above the water. This ball is found to hit the water with certain velocity before sinking into water with certain constant velocity. The ball is found to reach the bottom at the end of 3s. We are supposed to find the depth of the lake with the given information.
As the ball is dropped, the initial velocity would be,
$u=0$
From Newton’s equation of motion we have,
$s=ut+\dfrac{1}{2}g{{t}^{2}}$
$\Rightarrow s=0\times t+\dfrac{1}{2}g{{t}^{2}}$
But we know the height to be,
$s=5$
$\Rightarrow 5=\dfrac{1}{2}\times 10\times {{t}^{2}}$
$\therefore t=1s$
Therefore, we found that the time taken by the lead ball to hit the water would be 1s.
Now we could find the velocity with which the ball hit the water using,
${{v}^{2}}={{u}^{2}}+2gs$
$\Rightarrow {{v}^{2}}=0+2\times 10\times 5$
$\therefore v=\sqrt{100}=10m{{s}^{-1}}$
We are given that the lead ball takes 3s to reach the bottom of the lake after dropping, so the time taken t’ to cover the depth d of the water would be,
$t'=3-t=3-2=2s$
Let the constant velocity with which the ball covers the depth be the velocity with which it hit the water, then,
$d=v\times t'$
$\Rightarrow d=10\times 2$
$\therefore d=20m$
Therefore, we found the depth of the lake to be 20m, i.e., option D is correct.
Note:
As the ball is under free fall after being dropped, we could say that the only force on the ball will be that due gravity. So the acceleration of the ball would be that due to gravity. We have also made an assumption that the ball will have the same velocity that it had when it hit the surface of the lake.
Complete Step by step solution:
In the question, we are told that a lead ball is dropped into the lake from a height of 5m from a diving board above the water. This ball is found to hit the water with certain velocity before sinking into water with certain constant velocity. The ball is found to reach the bottom at the end of 3s. We are supposed to find the depth of the lake with the given information.
As the ball is dropped, the initial velocity would be,
$u=0$
From Newton’s equation of motion we have,
$s=ut+\dfrac{1}{2}g{{t}^{2}}$
$\Rightarrow s=0\times t+\dfrac{1}{2}g{{t}^{2}}$
But we know the height to be,
$s=5$
$\Rightarrow 5=\dfrac{1}{2}\times 10\times {{t}^{2}}$
$\therefore t=1s$
Therefore, we found that the time taken by the lead ball to hit the water would be 1s.
Now we could find the velocity with which the ball hit the water using,
${{v}^{2}}={{u}^{2}}+2gs$
$\Rightarrow {{v}^{2}}=0+2\times 10\times 5$
$\therefore v=\sqrt{100}=10m{{s}^{-1}}$
We are given that the lead ball takes 3s to reach the bottom of the lake after dropping, so the time taken t’ to cover the depth d of the water would be,
$t'=3-t=3-2=2s$
Let the constant velocity with which the ball covers the depth be the velocity with which it hit the water, then,
$d=v\times t'$
$\Rightarrow d=10\times 2$
$\therefore d=20m$
Therefore, we found the depth of the lake to be 20m, i.e., option D is correct.
Note:
As the ball is under free fall after being dropped, we could say that the only force on the ball will be that due gravity. So the acceleration of the ball would be that due to gravity. We have also made an assumption that the ball will have the same velocity that it had when it hit the surface of the lake.
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