Question

A lead ball is dropped into a lake from a diving board 5 meters (=h) above the water. It hits the water with a certain velocity and then sinks to the bottom with this same constant velocity. It reaches the bottom 5.0 sec (=t) after it is dropped. a. How deep (D) is the lake?

Answer 1

Answer:

$t_1 = \sqrt{\frac{2* (-5m)}{-9.8 m/s^2}}=1.01 s$

$v_f = 0 -9.8 m/s^2 (1.01s) =-9.90 m/s$

$D = vt$

$|D|=|-9.9 m/s *3.99 s|= |-39.49 m|= 39.49 m$

Explanation:

For this case we have the initial height given:

$y_i = 5 m$ above the water

We know that the total time from the initial point and the end point it's $t_{tot}= 5 s$

So we can use this kinematic formula in order to find the time to travel from the original height to the surface of the water:

$y_f = y_i + v_i t + \frac{1}{2}a t^2_1$

We assume that the initial velocity is 0, the final height since it's at the surface so it would be 0m, the acceleration for this case would be the gravity and negative since it's acting downward, replacing we have this:

$0 = 5m -\frac{1}{2}(9.8 m/s^2)t^2_1$

And solving for t1 we got:

$t_1 = \sqrt{\frac{2* (-5m)}{-9.8 m/s^2}}=1.01 s$

Now we can find the velocity at the surface of the water like this:

$v_f = v_i - gt$

$v_f = 0 -9.8 m/s^2 (1.01s) =-9.90 m/s$

Since we know that the total time from the begin and the end it's 5 s and from the calculation before we know that the time to reach the surface of the water is 1.01 s, then the time between the surface of the water and the final depth of the water is 5-1.01= 3.99 s

And then we can find the depth of the lake (D) using the formula for constant velocity like this:

$D = vt$

$|D|=|-9.9 m/s *3.99 s|= |-39.49 m|= 39.49 m$