$A$ man is standing on top of a building $100\, m$ high. He throws two balls vertically,one at $t = 0$ and the other after a time interval $\Delta t$ (less than $2\, s$). The second ball is thrown at a velocity half that of the first. The vertical gap between the first and second ball is $15\, m$ at $t = 2\, s$. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throws.

(A) Let the velocities of the two balls be $v_1$ and $v_2$ respectively.
Given $v_2 = v$,then $v_1 = 2v$.
Since the vertical gap between the balls remains constant,both balls must be moving with the same velocity at any given time $t > \Delta t$. This implies that the balls have reached their maximum heights and are falling,or they are at the same stage of motion.
However,the condition that the gap remains constant $15\, m$ implies that the difference in their displacements is constant,which occurs when they are both in free fall under gravity after being thrown upwards.
Let $y_1$ and $y_2$ be the displacements from the top of the building: $y_1 = v_1 t - \frac{1}{2} g t^2$ and $y_2 = v_2 (t - \Delta t) - \frac{1}{2} g (t - \Delta t)^2$.
The gap $y_1 - y_2 = 15$. Substituting $v_1 = 2v$ and $v_2 = v$:
$(2v t - \frac{1}{2} g t^2) - (v(t - \Delta t) - \frac{1}{2} g (t - \Delta t)^2) = 15$.
Simplifying,$2vt - \frac{1}{2}gt^2 - vt + v\Delta t + \frac{1}{2}g(t^2 - 2t\Delta t + \Delta t^2) = 15$.
$vt + v\Delta t - g t \Delta t + \frac{1}{2}g \Delta t^2 = 15$.
For the gap to be constant,the coefficient of $t$ must be zero: $v - g \Delta t = 0 \implies v = g \Delta t$.
Substituting $v = 10 \Delta t$ into the constant term: $10 \Delta t^2 + 10 \Delta t^2 - 10 \Delta t^2 + 5 \Delta t^2 = 15 \implies 5 \Delta t^2 = 15 \implies \Delta t^2 = 3 \implies \Delta t = \sqrt{3} \approx 1.732\, s$.
Then $v = 10 \times 1.732 = 17.32\, m/s$. Thus $v_1 = 34.64\, m/s$ and $v_2 = 17.32\, m/s$.