Question

A lead ball is dropped into a lake from a diving board 12.0 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

.

Applying conservation of energy between the points A and B :

At point A , Kinetic energy of the ball = 0 (as ball was released from the rest)

Potential energy of the ball = mgh (where, g = 32.174 ft/s² is acceleration due to gravity, and h = 12 ft is height of the ball and m is mass of the ball)

At point B : Kinetic energy of the ball = 0.5mv² (where, v is velocity of the ball at the point B )

Potential energy of the ball = 0 ( as height = 0)

Applying conservation of energy, we get

  

or,  

or,  

or,  

or,

Hence, the ball hits the water surface with a velocity of 27.788 ft /s .

Now, we know the equation of motion,

  

Where, v is final velocity (velocity of ball at B)

u is initial velocity (velocity of ball at A )

a is the acceleration of the ball

t is time taken in the motion

For the given problem,

So,   

  

or,

Hence, time taken by ball to reach from A to B is 0.864 s .

Motion of ball between the points B and C :

According to the question, the ball inside the water moves with constant velocity. And it takes 5.30 s to reach to the bottom after it dropped from the board.

So, time taken by ball in the journey from B to C = 5.3 - 0.864 = 4.436 s

From the relation of velocity , distance and time  , we can write

  

Hence, the lake is 123.277 ft (nearly 123.3 ft) deep.

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