A ball is thrown from a point with a speed v0 at an angle of projection . From the same point and at this same instant, a person starts running with a constant speed v0/2 to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection?
Answers
Answered by
44
Man will catch the ball if the horizontal component of velocity becomes equal to the constant speed
of man i.e. θ=600
Approved
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
Answered by
83
Answer:
60°
Explanation:
Solution is in the attachment provided below.
Hope it helps!
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