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?

Man will catch the ball if the horizontal component of velocity becomes equal to the constant speed
of man i.e. θ=60⁰

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 = 0^{0 ...............................which is not possible}

...therefore  2 theta= 180-theta.........by trigonometry

                    3 theta = 180           =====>theta=60⁰