Graph Interpretation

Question 24: Height of the Tower

Phase 1 (O to A, $0 < t < 1$): Only the first stone is falling. Separation $s = x_1 = \frac{1}{2}gt^2$. This is parabolic.
Phase 2 (A to B, $1 < t < 3$): At $t=1$, the second stone is dropped. Both are in free fall. $$ s(t) = x_1(t) – x_2(t) = \frac{1}{2}gt^2 – \frac{1}{2}g(t-1)^2 $$ Expanding this: $s(t) = \frac{1}{2}g [t^2 – (t^2 – 2t + 1)] = \frac{1}{2}g(2t – 1)$. This equation is linear in $t$, which corresponds to the straight line AB in the graph.

Point B ($t=3$): The graph changes shape again at $t=3$. This indicates the relative motion has changed: the first stone has hit the ground. Therefore, the total time of flight for the first stone is $T = 3$ seconds.

Height Calculation: $$ H = \frac{1}{2} g T^2 = \frac{1}{2} (10) (3)^2 = 5(9) = 45 \text{ m} $$

---

Question 25: State of Second Stone

We established that Stone 1 hits the ground at $t=3$ seconds. We need the state of Stone 2 at this exact instant.

• Time of fall for Stone 2: It was dropped at $t=1$. So at $t=3$, it has fallen for $\Delta t = 3 – 1 = 2$ seconds.
• Velocity of Stone 2: $v = g(\Delta t) = 10 \times 2 = 20 \text{ m/s}$.
• Distance Fallen: $y = \frac{1}{2} g (\Delta t)^2 = 5(2)^2 = 20 \text{ m}$.
• Height above ground: $h = H – y = 45 – 20 = 25 \text{ m}$.