Question

A lead ball is dropped into a lake from a diving board 5.60 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 3.50 after it is released.

How deep is the lake?

Answer 1

We are told that the velocity in the water is the same as the velocity with which the ball hit the water, so the first thing we need to do now is find the velocity with which it hit the lake. This will be easiest if you use conservation of energy. Set up the equation as follows:

Initial Energy = Final Energy
Initial Kinetic + Initial Potential = Final Kinetic + Final Potential
or
K0 + U0 = Kf + Uf

The ball was initially at rest on the diving board, so K0=0, and gravitational potential energy is given by U=mgh, so the initial potential is U=mg(5.6m) , the final potential energy is zero because the ball is at ground level, or h=0. Also remember that Kf = (1/2) m v^2 , now plug in these values to the conservation equation:

0 + mg(5.6m) = (1/2) m v^2 + 0
mg(5.6 m) = (1/2) m v^2

divide the mass from both sides

(5.6m) g = (1/2) v^2

Now multiply both sides by 2

(11.2 m) g = v^2

Now, plug in 9.81 m/s^2 for g and then take the square root of both sides

v = ~ 10.5 m/s

So, the ball is traveling at 10.5 m/s while in the water. Now, the ball hits the bottom of the lake 3.5 seconds after released, so we now need to find out how long it took for the ball to fall from the diving board and hit the water, for this we will use the following kinematic equation:

xf= v0*t + (1/2)*a*t^2

xf is the distance, v0 is the initial velocity, a is acceleration, and t is time
xf=5.6 m
v0= 0 m/s
a= 9.81 m/s^2
Solve for t

5.6 = (1/2)*(9.81 m/s^2)*t^2

Simplify now to solve for t,

t = 1.07 s

So, the ball takes 1.07 s to hit the water, which means that the ball is then falling through the water at 10.5 m/s for the remaining 2.43 seconds, using the following formula we can find how far it fell in that time:

d = v*t

Plug in our known t and v

d = (10.5 m/s)*(2.43 s)

d = 25.5 m

So the lake is 25.5 meters deep

Hope this helped!

Answer 2

velocity with which ball hit the water = sqroot(2x9.8x5.60)=10.476 m/s apprx

Depth of lake= 10.476x 3.50s=36.668 m apprx