Question

A lead ball is dropped into a lake from a diving board 16.0 ft above the water. It hits the water with a certain velocity and then sinks to the bottom with this same constant velocity. If it reaches the bottom 5.30 s after it is dropped, how deep is the lake (in feet)?

Answer 1

138.46 ft

Step-by-step explanation:

When the ball is dropped until the moment it hits the water, the ball moves in a uniform acceleration motion. Therefore, the equation that describes the movement of the ball is:

`X = (1)/(2)*g*t^(2)  + V_(0) *t + x_0`

Where X is the distance that the ball has fallen at a time t.
`V_0` is the initial velocity, which is 0 ft/s as the ball was simply dropped.
`x_0` is the initial position, we will say that this value is 0 in the position where the ball was dropped for simplicity, and it increases as the ball is falling. Now, we replace x with 16 feets and solves for t:

`16 ft = (1)/(2) * 32.2 (ft)/(s^(2)) *t^(2) +0 (ft)/(s) * t + 0 ft`

`t = \sqrt{2* (16 ft)/(32.2 (ft)/(s^2))} = 1 s`

The velocity that the ball will have at the moment the ball that the ball hits the water will be:

`V = V_o+g*t=0(ft)/(s)+32.2(ft)/(s^2)*1s =32.2(ft)/(s)`

The time that will take the ball to reach the bottom from the top of the lake will be t = 5.3s - 1s = 4.3s. And as the ball will travel with constant velocity equal to 32.2 ft/s^2, the depth of the lake will be:

`d = v*t = 32.2(ft)/(s)  * 4.3s = 138.46 ft`