Question

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

• 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 1

Answer: (C)

$x_{2} - x_{1} = 15t$

Answer 2

As $x = x_{0} + ut + \frac{1}{2}at^{2}$

For first stone,

$x = x_{1},x_{0} = 200m,u = 15ms^{- 1}$

$a = - g = - 10ms^{- 2}$

$\therefore \mathrm { x } _ { 1 } = 200 + 15 \mathrm { t } + \frac { 1 } { 2 } \times ( - 10 ) \times \mathrm { t } ^ { 2 }$ $x_{1} = 200 + 15t - 5t^{2}$ ……. (i)

When this hits the ground, $x_{1} = 0$

$\therefore 200 + 15t - 5t^{2} = 0$

$t^{2} - 3t - 40 = 0$

$t^{2} - 8t + 5t - 40 = 0$

$t(t - 8) + 5(t - 8) = 0$

$\Rightarrow t = 8sor - 5s$

Since the stone was projected at t = 0, Hence, negative time has no meaning in this case.

For the second stone,

$x = x_{2},u = 30ms^{- 1},a = - g = - 10ms^{- 2}$

$x_{0} = 200m$

$\therefore x_{2} = 200 + 30t - 5t^{2}$ ……. (ii)

When this stone hits the ground, $x_{2} = 0$

$\therefore - 5t^{2} + 30t + 200 = 0$

$t^{2} - 6t - 40 = 0$

$t^{2} - 10t + 4t - 40 = 0$

$t(t - 10) + 4(t - 10) = 0$

$(t + 4)(t - 10) = 0$

$\Rightarrow$t = - 4s,or 10s

$\because$ t = - 4 s is meaningless

Therefore, t = 10s

Subtracting (ii) from (i) we get

$x_{2} - x_{1} = (200 + 30t - 5t^{2}) - (200 + 15t - 5t^{2}) = 15t$