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

Answer

`3m ; 45 m`

Explanation

Distance travelled by `2nd`
Particle in `2 s = 0.5 xx 2 = 1m`
Horizontal range `= 1 + 1 + 1 = 3 m`
Flight time `= 4 + 2 = 6 s`
`6 = (2 u sin theta)/(g) u sin theta = 30`
`H = (u^2 sin^2 theta)/(2 g) = (900)/( 2 xx 10) = 45`
Particle will strike the ground after `2 s`.
.