Question

Two particles were projected one by one with same initial velocity from the same point on the level ground. They follow the same parabolic trajectory and are found to be in the same horizontal level, separated by a distance of 1m, 2s after the second particle was projected. Assume that the horizontal component of their velocities is 0.5 ms-1. Find the horizontal range of the parabolic path

Answer 1

Explanation:

Distance travelled by 2nd particle in $$2\,s$$ = $$0.5\times 2=1m$$

Horizontal range $$=1+1+1=3m$$

Flight time $$= 4+2=6s$$

$$6=\dfrac{2u\sin \theta}{g} u\sin \theta =30$$

$$H=\dfrac{u^2\sin^2\theta}{2g}=\dfrac{900}{2\times 10}=45$$