There is a projectile which is launched at point "o" and land at "B".It was thrown with 20m/s with angle of 53 degree.Take g=10 m/s^2 sin theta=0.8 and cos theta=0.6
Find the time at which magnitude of x component of velocity of projectile is twice the magnitude of y component of velocity.
v sub x =u sub x +a sub x t
u sub x=initial velocity in x-direction i.e u cos theta =16 m/s
v sub x =velocity in x direction at time t.
v sub y =velocity in y direction at time t
v sub y =u sin theta +(- 10 t)
The Attempt at a Solution
v sub x=2 (u sin theta +(- 10 t)
but v sub x=u cos theta (as acceleration in x i.e horizontal direction is zero.
so,u cos theta =2 (u sin theta - 10 t)
putting given values give
But in book two answer are given
and t=2.2 sec.
How I can get t=2.2 s ?