Question

An object is thrown vertically upward from the ground passes the height 5m twice in an interwal of 10s what is the time of flight.

Answer 1

Time of ascent is always equal to time of descent.

Hence,time of flight=time of ascent+time of descent,

Total time of flight=2*(time of descent)

Now after reaching maximum height it took 5 sec(10/2) for object to reach a height of 5mts from the ground…

It’s velocity at that point=0+9.81*5(u+at) in downward direction

Velocity=49.5m/sec

Now let’s forget all the above steps …

A body with initial velocity 49.5 m/sec and acceleration 9.81m/sec^2,has to ravel 5mts,

Time taken=

5=49.5t+9.81t^2(s=ut+1/2at^2)

t=0.09sec

Time of descent=5+0.09=5.09 sec

Time of flight=2*5.09=10.18sec