A lead ball is dropped into a lake from a diving board 5.20 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 4.50 s after it is released. How deep is the lake?

35.047 m

Explanation:

The time it takes the lead ball to reach the surface of the water is

s = ut+gt²/2 Equation 1

Where t = time it takes the lead ball to reach the surface of water, u = initial velocity of the lead ball, g = acceleration due to gravity, s = heigth.

From the question,

Given: s = 5.20 m, u = 0 m/s (dropped from a height)

Constant: g = 9.8 m/s²

5.2 = 0+9.8t²/2

t² = (5.2×2)/9.8

t² = 10.4/9.8

t² = 1.06

t = √(1.06)

t = 1.03 s

Hence, time taken for the lead ball to reach the bottom of the lake is

t' = 4.5-1.03

t' = 3.47 seconds

v² = u²+2gs Equation 2

Where v = final velocity of the lead ball

Substitute into equation 2

v² = 0+2(9.8)(5.2)

v² = 101.92

v = √(101.92)

v = 10.1 m/s

Therefore, depth of the lake is

D = vt'

D = 10.1(3.47)

D = 35.047 m

increase masses, decrease separation