# Calculate the time of flight of a body which is thrown up to a height of 5 m from the ground ?

Feb 9, 2018

If the body is thrown up with a velocity of $u$,at the highest point (a height of $h$) of its motion,it will loose its upward velocity,and to reach that height if it takes time $t$,

we can say, $0 = u - g t$ (using $v = u - g t$)

Or, $t = \frac{u}{g}$ or, $t = \frac{u}{g}$

and,

${0}^{2} = {u}^{2} - 2 g h$ (using ${v}^{2} = {u}^{2} - 2 g h$)

Or, ${u}^{2} = 2 g h$

or, $u = \sqrt{2 g h}$

Now, when it comes down,if it achieves a velocity of $v$ while striking the ground,we can say, ${v}^{2} = 2 g h$ or, v=sqrt(2gh)(where, h# is the maximum height reached by the body)

CLEARLY THE BODY RETURNS TO GROUND WITH THE SAME VELOCITY WITH WHICH IT WAS PROJECTED AS WE GOT BOTH $v = u = \sqrt{2 g h}$

Again,if during coming down,it takes a time of $t '$,we can say $v = g t '$ or. $t ' = \frac{v}{g}$

So,time of flight ($T$) = $t + t ' = \frac{u}{g} + \frac{v}{g} = \left(\frac{1}{g}\right) \left(\sqrt{2 g h} + \sqrt{2 g h}\right) = \frac{1}{g} \cdot 2 \cdot \sqrt{2 g h}$

Given, $h = 5 m$

So,putting the value we get, $T = 2 s$

REMEMBER-when a body is thrown up with velocity $u$,total time of flight is expressed as $\frac{2 u}{g}$