a projectile is thrown with speed 40 m/s at an angle q from horizontal . it is found that projectile is at same height at 1 seconod and 3 second. what is the angle of projection
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It is at same height at time 1s and 3s.
Therefore, time taken to reach the maximum height will be 2s.
Final speed at maximum height will be zero.
So,v = usinq - at ( Negative sign because
it is moving upwards.)
usinq = at (Since,v=0)
40*sinq = 10*2
sinq = 1/2
q = 30.
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