Question

A lead ball is dropped into a lake from a diving board 6.10 mm 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 ss after it is released. How deep is the lake?

Answer 1

Answer: d = 1.54m

The depth of the lake is 1.54m

Explanation:

The final velocity of the ball just before it hit the water can be derived using the equation below;

v^2 = u^2 + 2as ......1

Where ;

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled.

Since the initial velocity is zero, and the acceleration is due to gravity, the equation becomes:

v^2 = 2gs

v = √2gs ......2

g = 9.8m/s^2

s = 6.10mm = 0.0061m

substituting into equation 2

v = √(2 × 9.8× 0.0061)

v = 0.346m/s

The time taken for the ball to hit water from the time of release can be given as:

d = ut + 0.5gt^2

Since u = 0

d = 0.5gt^2

Making t the subject of formula.

t = √(2d/g)

t = √( 2×0.0061/9.8)

t = 0.035s

The time taken for the ball to reach the bottom of the lake from the when it hits water is:

t2 = 4.5s - 0.035s = 4.465s

And since the ball falls for 4.465s to the bottom of the lake at the same velocity as v = 0.346m/s. The depth of the lake can be calculated as;

depth d = velocity × time = 0.346m/s × 4.465s

d = 1.54m

The depth of the lake is 1.54m

Answer 2

Answer:

D=1.54489 m

Explanation:

Given data

S=6.10 mm= 0.0061 m

To find

Depth of lake

Solution

To find the depth of lake first we need to find the initial time ball takes to hit the water.To get the value of time use below equation

$S=v_{1}t+(1/2)gt^{2} \\ 0.0061m=(0m/s)t+(1/2)(9.8m/s^{2} )t^{2}\\ t^{2}=\frac{0.0061m}{4.9m/s^{2} }\\ t=\sqrt{1.245*10^{-3} }\\ t=0.035s$

So ball takes 0.035sec to hit the water

As we have found time Now we need to find the final velocity of ball when it enters the lake.So final velocity is given as

$v_{f}=v_{i}+gt\\v_{f}=0+(9.8m/s^{2} )(0.035s)\\ v_{f}=0.346m/s$

Since there are (4.50-0.035) seconds left for (ball) it to reach the bottom of the lake

So the depth of lake given as:

$D=|vt|\\D=|0.346m/s*4.465s|\\D=1.54489m$