Q.A man is standing on top of a building 100 m high. He throws two balls vertically, one at $t = 0$ and other after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between 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 throw.

Concept: Motion under gravity (1D) — constant vertical gap between two thrown balls requires equal velocities at that instant.

Step 1. Let ball 1 be thrown with speed $u$ upward at $t=0$, and ball 2 with speed $\dfrac{u}{2}$ at $t=\tau$ (both later fall under $g=10\ \text{m/s}^2$).

Step 2. A constant gap after $t=2\ \text{s}$ means the two balls have equal velocity at $t=2\ \text{s}$:

Setting $v_1(2)=v_2(2)$: $u - 20 = \dfrac{u}{2} - 20 + 10\tau \Rightarrow \dfrac{u}{2} = 10\tau \Rightarrow u = 20\tau$.

Step 3. Positions at $t=2\ \text{s}$ (measured from the throw point, upward positive):

$y_1(2) = 2u - \tfrac12 g(2)^2 = 2u - 20, \qquad y_2(2) = \tfrac{u}{2}(2-\tau) - \tfrac12 g(2-\tau)^2$ …

Why the gap becomes constant

When two objects move under gravity alone, their separation changes unless they share the same velocity. Think of it this way: if ball 1 is moving faster upward (or slower downward) than ball 2, the gap between them grows; if slower upward (or faster downward), the gap shrinks. A constant gap means both balls are moving at exactly the same speed at every instant after $t = 2\,\text{s}$.

Since both balls experience the same acceleration $g = 10\,\text{m/s}^2$ downward, their velocities at time $t$ are:

where $u_1$ and $u_2$ are the initial speeds, and ball 2 is thrown $\Delta t$ seconds after ball 1.

For the gap to remain constant from $t = 2\,\text{s}$ onward, we need $v_1(2) = v_2(2)$:

We are also told $u_2 = u_1/2$, so:

Finding the separation at $t = 2\,\text{s}$

The position of each ball at time $t$ (measured downward from the top of the building as positive) is:

The vertical gap (ball 1 ahead, so $s_1 > s_2$) at $t = 2\,\text{s}$ is:

Substitute the positions:

Solving the system

From equation (2), $u_1 = 20\Delta t$ and $u_2 = u_1/2 = 10\Delta t$.

Substitute into equation (3):

$2(20\Delta t) - 10\Delta t(2 - \Delta t) + 5(2 - \Delta t)^2 = 35$ …

Concept: Constant Relative Velocity Under Equal Accelerations — Evaluate the Equal-Velocity Condition at the Natural Instant, Not at $t=2\,\text{s}$

Method: The Relative-Velocity-Invariant Shortcut — Both Conditions Read Off at the Moment Ball 2 is Thrown, Never Needing the Given $t=2\,\text{s}$ in the Algebra

The stored answer evaluates $v_1(2)=v_2(2)$ and separately builds full position expressions at $t=2\ \text{s}$ (with $(2-\tau)$ terms) to use the $15\ \text{m}$ gap. This method uses one structural fact — since both balls share the identical acceleration $-g$ once both are in flight, their relative velocity is constant for the entire time they're both airborne, not just from $t=2\ \text{s}$ onward — to derive both required equations at the single most natural moment: $t=\tau$, the instant ball 2 is thrown, when ball 2's own position and velocity are both known immediately from its initial conditions.

Steps

1. State the key structural fact. For $t\ge\tau$ (both balls in flight), $v_1(t)=u_1-gt$ and $v_2(t)=u_2-g(t-\tau)$, so their difference is:

This expression has no $t$ in it at all — the relative velocity is automatically the same constant at every instant $t\ge\tau$, simply because both balls share the same acceleration. "The gap remains constant" (i.e. $v_1-v_2=0$) is therefore either true for every $t\ge\tau$ at once, or false for all of them — there is nothing special about the given $t=2\ \text{s}$ in this condition.

2. So the equal-velocity condition can be evaluated at the SIMPLEST possible instant — $t=\tau$ itself — instead of at $t=2\ \text{s}$. At $t=\tau$: $v_1(\tau) = u_1-g\tau$, and $v_2(\tau) = u_2$ (ball 2 has just been thrown, so $(t-\tau)=0$ and its velocity is exactly its initial speed). Setting them equal:

reached without ever substituting $t=2$.

3. Use the given relation $u_2 = u_1/2$ directly in $(\star)$:

4. Get the second equation the same way — evaluate the GAP itself at $t=\tau$, not at $t=2\ \text{s}$. At the instant ball 2 is thrown, its position is (by definition) exactly where it started: $y_2(\tau)=0$. So the gap at that instant is simply

And because the gap is constant for all $t\ge\tau$ (established in Step 1, once $(\star)$ holds), this value at $t=\tau$ must equal the same constant $15\ \text{m}$ given at $t=2\ \text{s}$ — the problem's "$15\ \text{m}$ at $t=2\ \text{s}$" and "the gap at the moment of the second throw" are one and the same number, precisely because the gap never changes once $(\star)$ is satisfied:

5. Solve the system. Substitute $u_1=20\tau$ from Step 3:

Then $u_1=20(1)=20\ \text{m/s}$, and $u_2=u_1/2=10\ \text{m/s}$. …