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

`y(t_(2))=19.7m`

Step-by-step 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}=√(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`