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=[V^{2}sin2(theta)]/gtime taken by the ball to reach the ground=T=2Vsin(theta)/gnote;] 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] = '''' '''' ( V^{2}sin 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