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

Answer:

28.8 meters

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 $\frac{m}{2} *v^{2} = m*g*h$

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$\frac{m}{s^{2} }$.

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 ($K=\frac{m}{2} *v^{2}$. And all that potential was converted to kinetic energy.

Now, from $\frac{m}{2} *v^{2} = m*g*h$ we can deduce that $v=\sqrt{2*g*h}$

Therefore v=9.6$\frac{m}{s}$

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$\frac{m}{s}$*3$s$=28.8$m$

Which is our answer.

Answer 2

Answer:

Depth of lake is 19.72 m      

Explanation:

We have equation of motion s = ut + 0.5at², where s is displacement, u is initial speed , a is acceleration and t is the time.

First let us find the time for it takes to reach the water surface,

           s = 5 m

           u = 0m/s

           a = 9.8m/s²

 Substituting,

                   5 = 0 x t + 0.5 x 9.8t²  

                    t = 1.01 seconds

Now we have equation of motion v = u + at, v is the final velocity

Let us find velocity of ball at the surface of water

              v = 0 + 9.81 x 1.01

              v =9.91 m/s

So the velocity of ball in water is 9.91 m/s

Total time = 3 s

Time in which ball is in water = 3 - 1.01 = 1.99 s

We have s = ut + 0.5at², we need to find s

             u = 9.91 m/s

             t = 1.99 s

             a = 0 m/s²

 Substituting,

                   s = 9.91 x 1.99 + 0.5 x 0 x 1.99²  

                   s = 19.72 m

Depth of lake is 19.72 m