Question

Mukesh drops a ball of 200 g from a tower of height
20 m. What will be its kinetic energy at the height of
5 m ?​

Answer 1

Answer:

Explanation:

K.E  = $\frac{1}{2}$ M x $V^{2}$

Since V = $\sqrt{2gh}$

and h = 20 - 5 = 15m

K.E = 1/2  X 0.2 kg x $(\sqrt{2*10*15})^{2}$

     =  0.1 x 300

    =   30 kg m/s^2

Answer 2

Given

A ball of weight 200g drops from height 20m

To find

Kinetic energy at the height of 5 m

Solution

We can simply solve this mathematical problem using the following mathematical process.

We have to find the kinetic energy of a ball at the height of 5 m  whose mass is 200 g and it is thrown from a height of 20 m.

Let the initial height be H and final height be h.

m = 200 g = $\frac{200}{1000}$  kg = 0.2 kg

We know that when the ball comes down its potential energy gets converted into its kinetic energy.

Kinetic energy = change in potential energy

K.E = mgH - mgh

K.E = mg(H-h)

= 0.2 ×10(20-5)

= 30 joules

Hence the kinetic energy of the ball at a height of 5 m is 30 joules.