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

Answer:

138.46 ft

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:

$X = \frac{1}{2}*g*t^{2} + V_{0} *t + x_0$

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:

$16 ft = \frac{1}{2} * 32.2 \frac{ft}{s^{2}} *t^{2} +0 \frac{ft}{s} * t + 0 ft$

$t = \sqrt{2* \frac{16 ft}{32.2 \frac{ft}{s^2}}} = 1 s$

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

$V = V_o+g*t=0\frac{ft}{s}+32.2\frac{ft}{s^2}*1s =32.2\frac{ft}{s}$

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:

$d = v*t = 32.2\frac{ft}{s} * 4.3s = 138.46 ft$