Question

Two balloons are simultaneously released from two buildings A and B. Balloon from A rises with constant velocity $10 \ ms^{-1}$, while the other one rises with constant velocity of $20 \ ms^{-1}$. Due to wind the balloons gather horizontal velocity $V_x = 0.5 \ y$, where 'y' is the height from the point of release. The buildings are at a distance of 250 m & after some time 't' the balloons collide. Choose the correct alternatives.

• A. t = 5 sec.
• B. difference in height of buildings is 100 m
• C. difference in height of buildings is 500 m
• D. t = 10 sec

Answer 1

Answer: (B), (D)

(B) and (D)

Answer 2

Let $H_A$ and $H_B$ be the initial heights of balloons A and B from the ground. Vertical positions: $y_A(t) = H_A + 10t$, $y_B(t) = H_B + 20t$. Horizontal velocities: $V_{xA}(t) = 0.5 y_A(t)$, $V_{xB}(t) = 0.5 y_B(t)$. Horizontal positions: $x_A(t) = \int_0^t 0.5 (H_A + 10\tau) d\tau = 0.5 (H_A t + 5t^2)$. $x_B(t) = 250 + \int_0^t 0.5 (H_B + 20\tau) d\tau = 250 + 0.5 (H_B t + 10t^2)$. For collision: $y_A(t) = y_B(t) \implies H_A + 10t = H_B + 20t \implies H_A - H_B = 10t$. Also, $x_A(t) = x_B(t) \implies 0.5 (H_A t + 5t^2) = 250 + 0.5 (H_B t + 10t^2)$. $H_A t + 5t^2 = 500 + H_B t + 10t^2 \implies (H_A - H_B)t - 5t^2 = 500$. Substitute $H_A - H_B = 10t$: $(10t)t - 5t^2 = 500 \implies 10t^2 - 5t^2 = 500 \implies 5t^2 = 500 \implies t^2 = 100 \implies t = 10$ s. Then, $H_A - H_B = 10t = 10(10) = 100$ m.