Question

A stone is released from the top of a tower of height 20m. Calculate the velocity with which the stone hits the ground. Take the value of 'g' as 10ms 2.​

Answer 1

Answer:

Velocity of stone which hits the ground is 20 m/s.

Explanation:

Given :

u = 0 m/s

s = 20 m

g = 10 m/s^2

To find

v = ?

According to the question

By the Newton's 3rd law of equation of motion,

We have,

${v}^{2} = \: {u}^{2} + 2gs \\ {v }^{2} - {u}^{2} = 2gs \\ {v}^{2} - {0}^{2} = 2 \times 10 \times 20 \\ {v}^{2} = \: 400 \\ v \: = \sqrt{400} \\ v = 20 \: metre \: per \: second$

Here,

v is the velocity of stone at which , stone hits the ground,

u is the initial velocity of stone,

s is the height of the tower,

g is the acceleration due to gravity.

If body is released , then we take initial velocity is 0 m/s.

Velocity of stone which hits the ground is 20 m/s.