Question

A lead ball is dropped into a lake from a diving board 5.0 meters 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 seconds after it is released. How deep is the lake?

Answer 1

The depth of the lake can be calculated by determining the time it takes for the ball to reach the bottom. Using the equation d = 1/2gt^2, where d is the distance fallen, g is the acceleration due to gravity, and t is the time taken, we can solve for t. The time taken to fall from the diving board is subtracted from the total time taken to sink to the bottom, and then the distance fallen from the diving board to the bottom of the lake is calculated using the equation d = 1/2gt^2. The depth of the lake is found to be 19.42 meters.

Step-by-step explanation:

To determine the depth of the lake, we first need to calculate how long it takes for the ball to reach the bottom. Since the ball sinks with a constant velocity, the time it takes to reach the bottom is the same as the time it takes to fall from the diving board. Using the equation d = 1/2gt^2, where d is the distance fallen, g is the acceleration due to gravity (9.8 m/s^2), and t is the time taken, we can solve for t: 5 = 1/2 * 9.8 * t^2. Rearranging the equation, we have t^2 = 5 / (1/2 * 9.8), which simplifies to t^2 = 1.02. Taking the square root of both sides, we find t = 1.01 s. Since the ball reaches the bottom after 3.0 s, we subtract the time taken to fall from the diving board to get the time taken to sink to the bottom: 3.0 s - 1.01 s = 1.99 s.

Next, we calculate the distance fallen from the diving board to the bottom of the lake using the equation d = 1/2gt^2. The time taken is 1.99 s, and the acceleration due to gravity is 9.8 m/s^2. Plugging in the values, we get d = 1/2 * 9.8 * (1.99)^2. Simplifying the equation, we find d = 19.42 m. Therefore, the depth of the lake is 19.42 meters.

Answer 2

28.8 meters

Step-by-step explanation:

We must first determine at which velocity the ball hits the water. To do so we will:

1) Assume no air resistance.

2) Use the Law of conservation of mechanical energy: E=K+P

Where

E is the mechanical energy (which is constant)

K is the kinetic energy.

P is the potential energy.

With this we have
`(m)/(2) *v^(2)  = m*g*h`

Where:

m is the balls's mass <- we will see that it cancels out and as such we don't need to know it.

v is the speed when it hits the water.

g is the gravitational constant (we will assume g=9.8
`(m)/(s^(2) )`.

h is the height from which the ball fell.

Because when we initially drop the ball, all its energy is potential (and
`P = - m*g*h`) and when it hits the water, all its energy is kinetic (
`K=(m)/(2) *v^(2)`. And all that potential was converted to kinetic energy.

Now, from
`(m)/(2) *v^(2)  = m*g*h` we can deduce that
`v=√(2*g*h)`

Therefore v=9.6
`(m)/(s)`

Now, to answer how deep is the lake we just need to multiply that speed by the time it took the ball to reach the bottom.

So D=9.6
`(m)/(s)`*3
`s`=28.8
`m`

Which is our answer.