Question

Two stones $A$ and $B$ are thrown with speed $5\sqrt{3}$ m/s from top of a tower of height 10 m at some interval of time. $A$ is thrown horizontally and $B$ is thrown at angle $30^\circ$ with vertically upward direction. They collide in air at point $P$. Choose correct option(s).

• A. The time interval between the instants of throwing $A$ and $B$ is $\sqrt{2}$ sec
• B. The time interval between the instants of throwing $A$ and $B$ is 1 sec
• C. Distance of point $P$ from foot of the tower is 10 m
• D. Distance of point $P$ from foot of the tower is $10\sqrt{3}$ m

Answer 1

Answer: (B), (C)

B, C

Answer 2

Let the origin be at the top of the tower, with the positive x-axis horizontal and the positive y-axis vertically upwards. The height of the tower is $H=10$ m and $g=10$ m/s². The initial speed for both stones is $v_0 = 5\sqrt{3}$ m/s.

For stone A (thrown horizontally): Initial velocity: $\vec{v}_{A0} = (5\sqrt{3}, 0)$ m/s. Position at time $t_A$: $x_A(t_A) = 5\sqrt{3} t_A$, $y_A(t_A) = -5 t_A^2$.

For stone B (thrown at $30^\circ$ with vertically upward direction, so $60^\circ$ with horizontal): Initial velocity: $v_{B0x} = 5\sqrt{3} \cos(60^\circ) = \frac{5\sqrt{3}}{2}$ m/s, $v_{B0y} = 5\sqrt{3} \sin(60^\circ) = \frac{15}{2}$ m/s. Position at time $t_B$: $x_B(t_B) = \frac{5\sqrt{3}}{2} t_B$, $y_B(t_B) = \frac{15}{2} t_B - 5 t_B^2$.

Let $\Delta t$ be the time interval. Assume B is thrown first. Collision occurs at time $T$ (from B's throw). Time for A: $t_A = T - \Delta t$. Time for B: $t_B = T$. Equating x-coordinates for collision: $5\sqrt{3}(T-\Delta t) = \frac{5\sqrt{3}}{2}T \implies T-\Delta t = \frac{T}{2} \implies T = 2\Delta t$. So, $t_A = \Delta t$ and $t_B = 2\Delta t$.

Equating y-coordinates: $y_A(t_A) = y_B(t_B)$ $-5 (\Delta t)^2 = \frac{15}{2} (2\Delta t) - 5 (2\Delta t)^2$ $-5 \Delta t^2 = 15 \Delta t - 20 \Delta t^2$ $15 \Delta t^2 - 15 \Delta t = 0$ $15 \Delta t (\Delta t - 1) = 0$. Since $\Delta t \neq 0$, $\Delta t = 1$ sec. Thus, Option B is correct.

Collision point P: $t_A = 1$ sec. $x_P = 5\sqrt{3} \times 1 = 5\sqrt{3}$ m. $y_P = -5 \times (1)^2 = -5$ m. The collision point P is at $(5\sqrt{3}, -5)$ relative to the top of the tower.

Distance of P from the foot of the tower (at $(0, -10)$): $d = \sqrt{(5\sqrt{3}-0)^2 + (-5 - (-10))^2} = \sqrt{(5\sqrt{3})^2 + (5)^2} = \sqrt{75 + 25} = \sqrt{100} = 10$ m. Thus, Option C is correct.

Options A and D are incorrect as $\Delta t = 1$ sec and $d=10$ m.