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:
[tex]X = \frac{1}{2}*g*t^{2} + V_{0} *t + x_0[/tex]
Where X is the distance that the ball has fallen at a time t. [tex]V_0[/tex] is the initial velocity, which is 0 ft/s as the ball was simply dropped. [tex]x_0[/tex] 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:
[tex]16 ft = \frac{1}{2} * 32.2 \frac{ft}{s^{2}} *t^{2} +0 \frac{ft}{s} * t + 0 ft[/tex]
[tex]t = \sqrt{2* \frac{16 ft}{32.2 \frac{ft}{s^2}}} = 1 s[/tex]
The velocity that the ball will have at the moment the ball that the ball hits the water will be:
[tex]V = V_o+g*t=0\frac{ft}{s}+32.2\frac{ft}{s^2}*1s =32.2\frac{ft}{s}[/tex]
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:
[tex]d = v*t = 32.2\frac{ft}{s} * 4.3s = 138.46 ft[/tex]
