Question

a stone is projected vertically upward from the top of Tower with velocity of 19.6 m/s from the top of a tower 20.4m height. find the greatest elevation, its velocity at the moment it strikes the ground.​

Answer 1

Explanation:

It is given that,

Initial velocity with which a stone is projected vertically upward from the top of a tower 20.4 m high.

Let h is the greatest elevation.

Using conservation of energy,

$mgh=\dfrac{1}{2}mu^2\\\\u=\dfrac{u^2}{2g}\\\\u=\dfrac{(19.6)^2}{2\times 9.8}\\\\u=19.6\ m/s$

Let v is the velocity at the moment it strikes the ground.​ Using the third equation of motion as :

$v^2-u^2=2ah$

Here, a = g

So,

$v^2=2ah+u^2\\\\v=\sqrt{2gh+u^2} \\\\v=\sqrt{2\times 9.8\times 20.4+(19.6)^2} \\\\v=28\ m/s$

Hence, this is the required solution.