Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of `15 m s^(-1)` and `30 m s^(-1)` respectively. The time variation of the relative position of the second stone will request to the first as shown in the figure. The equation of the linear part is

Options

A. `x_2 - x_1` = 50t
B. `x_2 - x_1` = 10t
C. `x_2 - x_1` = 15t
D. `x_2 - x_1` = 20t

Answer

C

Explanation

As `x = x_0 + ut + 1/2 at^(2)`
For first stone,
`x =x_1, x_0 = 200 m, u = 15 m s^(-1)`
`a = -g = 10 m s^(-2) therefore x_1 = 200 + 15t + 1/2 xx (-10) xx t^(2)`
`x_1 = 200 + 15t - 5t^(2)` …..(i)
When this hits the ground, `x_1 = 0`
`therefore 200 + 15t - 5t^(2) = 0`, on solving, we get
`implies` t = 8 s or `-5`s
Since the stone was projected at t = 0. Hence, negative time has no meaning in this case.
For the second stone.
`x = x_2, u = 30 m s^(-1), a = -g = -10 m s^(-2), x_0 = 200 m `
`therefore x_2 = 200 + 30t - 5t^(2)` .....(ii)
When this stone hits the ground, `x_2` = 0
`therefore` `-5t^(2) + 30t + 200 = 0`, on solving, we get
`implies t = - 4 s`, or 10 s
since `t = - 4` s is meanigless. Therefore, t = 10 s
Subtracting (i) from (ii) we get
`x_2 - x_1 = (200 + 30t - 5t^(2)) - (200 + 15t - 5t^(2)) = 15 t`