Question

A body is thrown verticaly upwards with velocity u find the greatest height h to which. how much it will rises

Answer 1

Given

A body is thrown upward with an initial velocity 'u'.

To Find

Maximum height of the ball .

Concepts

• At maximum height,velocity of the body becomes zero.Therefore,it starts falling down as it is accelerated downward the whole journey.

Now, the question can be solved using two method ,the first method uses energy conservation which states :-

• In the absence of any external or nonconservative forces, mechanical energy of a body is conserved.

=> $(K.E.\:+\:P.E.)_{initial}\:=\:{(K.E.\:+\:P.E.)}_{final}$

The other method uses three formulas :-

$v\:=u\:+\:at\\ \\ s\:=\dfrac{at^2}{2}\\ \\ v^2\:-\:u^2\:=\:2as$

Solutions

Using energy conservation :-

At the bottom point, P.E. is zero while K.E. is zero at top point.

Therefore ,

$\dfrac{mu^2}{2}\:=\:mgh \\ \\ \\ =>\: u^2\:=\:2gh \\ \\ \\ => \:u\:= \sqrt{2gh} \\ \\ \\ => \: \boxed{h\:=\: \dfrac{u^2}{2g}}$

Using formulas

=> v = u + at

=> 0 = u - gt

=> u = gt

=> $t\:=\: \dfrac{u}{g}$

This time is called time of ascent.It is the time to reach maximum height. This is also equal to time of descent .

=> Time of ascent = Time of descent.

Therefore, total time spent in air is 2t = $\dfrac{2u}{g}$

Now, h can be find out by using :-

$h\:=\:ut\:+\dfrac{at^2}{2} \\ \\ h\:=\: \dfrac{u^2}{g}\:-\dfrac{u^2}{2g} \\ \\ \: \boxed{h\:=\: \dfrac{u^2}{2g}}$