Question

57. ••• A lead ball is dropped into a lake from a diving board 5.0 m above the water. After entering 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 3.0 s after it is released. How deep is the lake?Physics for Scientists and Engineers

Answer 1

Answer:

$y(t_{2})=19.7m$

Explanation:

Kinematics equation for the lead ball, before enter into the water:

$v(t)=v_{o}+g*t$

$y(t)=y_{o}+v_{o}t+1/2*g*t^{2}$

$v_{o}=0m/s$    The initial velocity is zero

when the ball reaches the water at time, t1, y(t)=5.0m:

$y(t_{1})=1/2*g*t_{1}^{2}$

$t_{1}=\sqrt{2y(t_{1})/g}=\sqrt{2*5/9.81}=1.01s$

final velocity:

$v_{1}(t)=g*t_{1}=9.81*1.01=9.9m/s$

Kinematics equation for the lead ball, after it enters into the water:

Velocity constant, equal to final velocity with which it hit the water, v=v1=9.9m/s

$y(t)=vt$

If the ball reaches the bottom 3.0s after it is released. The ball is into the water a time, t2, t2=3.0-t1=3-1.01=1.99s

The depth of the lake is:

$y(t_{2})=vt_{2}=9.9*1.99=19.7m$