A ball is thrown from a point with a speed V at an angle of projection θ . From the same point and at the same instant, a person starts running with a constant speed V/2 to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection?
Ans) Horizontal distance covered by the ball= horizontal range= (hope u know the formula) R=[V2sin2(theta)]/g
time taken by the ball to reach the ground=T=2Vsin(theta)/g
note;] remember that time taken by the person to reach the spot and time taken
by the ball to reach the ground(same spot)..is THE SAME;
time taken by the person to reach the spot= time taken by the ball..
distance/speed (man) = 2v sin theta/g [ball]
range/speed[man] = '''' ''''
( V2sin 2 theta/g) / v/2 = '''' ''''
things cancel out in the equation and finally u get sin (theta) = sin 2 (theta)
this can happen if theta = 00 ...............................which is not possible
...therefore 2 theta= 180-theta.........by trigonometry
3 theta = 180 =====>theta=600