Angle of projection, Mechanical Engineering

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

Posted Date: 3/12/2013 2:40:36 AM | Location : United States







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