Q.
A projectile is fired at an angle of 30∘ to the horizontal
such that the vertical component of its initial velocity is 80ms−1. Its time of flight is T Its velocity at the moment t=4T, has a magnitude of nearly (Take g=10ms−2)
Vertical component of initial velocity, uy=usin30∘
or u=sin30∘uy=(1/2)80=160ms−1
Horizontal component of initial velocity, ux=ucos30∘=160×23=803ms−1
Time of flight, T=g2usinθ =102×160×sin30∘=16s ∴t=4T=416=4s
Let v be the velocity of the projectile at t=4T
Its horizontal and vertical components are given by vx=ux=803ms−1 vy=uy−gt=80−10×4=40ms−1
Its magnitude is given by v=vx2+vy2=(803)2+(40)2 =4012+1=4013≈144ms−1