Question

3- a lead ball is dropped into a lake from a diving board 12.0 m above the water. after entering to 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 5.0 s after it is released. how deep is the lake?

Explanation:

Step1: Calculate the time to fall to the water surface

The ball is in free - fall motion before entering the water. The height of the diving board above the water is $h_1 = 12.0$ m, and the acceleration due to gravity is $g=9.8$ m/s². Using the equation $h_1 = v_0t_1+\frac{1}{2}gt_1^{2}$, with $v_0 = 0$ m/s, we have $h_1=\frac{1}{2}gt_1^{2}$. So, $t_1=\sqrt{\frac{2h_1}{g}}$.
$t_1=\sqrt{\frac{2\times12.0}{9.8}}\approx1.56$ s.

Step2: Calculate the time to sink in the water

The total time from release to reaching the bottom is $t = 5.0$ s. The time to sink in the water is $t_2=t - t_1$.
$t_2=5.0 - 1.56 = 3.44$ s.

Step3: Calculate the velocity when hitting the water

Using the equation $v = v_0+gt_1$ (with $v_0 = 0$ m/s), we get $v = gt_1$.
$v=9.8\times1.56\approx15.3$ m/s.

Step4: Calculate the depth of the lake

Since the ball sinks in the water with a constant velocity $v$, and the time of sinking is $t_2$, the depth of the lake $h_2=v\times t_2$.
$h_2 = 15.3\times3.44\approx52.6$ m.

Answer:

The lake is approximately $52.6$ m deep.