A ball thrown from ground with speed u at an angle ϴ with the horizontal, at the same instant from the same position a boy runs along the ground with speed [3^ (1/2)/2]u to catch the ball. If the boy can catch the ball, the angle of projection ϴ is

A ball thrown from ground with speed u at an angle ϴ with the horizontal, at the same instant from the same position a boy runs along the ground with speed u[3^ (1/2)/2] to catch the ball. If the boy can catch the ball, the angle of projection ϴ is

solution. Horizontal range of the projectile = The distance travelled by the boy, to catch the ball

               Horizontal range of projectile= u^2sin2ϴ/g----------1

               Distance travelled by the boy= [3^ (1/2)/2]u.2usinϴ/g-------------2

Equating 1 and 2,

u^2sin2ϴ/g=[3^ (1/2)/2]u.2usinϴ/g

          cosϴ=3^ (1/2)/2

               ϴ=30 degree(Ans)