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

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Jane Share
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=u/g# or, #t=u/g#

and,

#0^2=u^2 -2gh# (using #v^2=u^2 -2gh#)

Or, #u^2 = 2gh#

or, #u = sqrt(2gh)#

Now, when it comes down,if it achieves a velocity of #v# while striking the ground,we can say, #v^2=2g 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(2gh)#

Again,if during coming down,it takes a time of #t'#,we can say #v=g t'# or. #t' = v/g#

So,time of flight (#T#) = #t+t' = u/g +v/g = (1/g )(sqrt(2gh) + sqrt(2gh)) = 1/g * 2* sqrt(2gh)#

Given, #h=5m#

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 #(2u)/g#

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