Solution to Question 12
Problem Analysis: A frog must jump over a pipe moving towards it. The trajectory touches the pipe “only at the top,” implying tangency at the vertex of the parabola.
Part 1: Time of Flight ($T$)
For the frog to pass over the pipe exactly at the top, the maximum height of the jump ($H$) must equal the diameter of the pipe ($2r$).
$$H = \frac{gT^2}{8}$$Substituting $H = 2r$:
$$2r = \frac{gT^2}{8} \implies T^2 = \frac{16r}{g} \implies T = 4\sqrt{\frac{r}{g}}$$This matches option (a).
Part 2: Range Condition ($R$)
The critical condition for “touching only at the top” involves the relative curvature of the trajectories. In the frame of reference of the moving pipe:
- The pipe is a stationary circle of radius $r$.
- The frog has a horizontal velocity relative to the pipe: $v_{rel} = u_x + v$, where $u_x = R/T$.
For the parabolic path not to intersect the circle (except at the touch point), the parabola must be “wider” than the circle. Mathematically, the Radius of Curvature ($\rho$) of the parabola at the vertex must be greater than or equal to the radius of the pipe.
$$\rho \ge r$$The radius of curvature of a projectile at maximum height is given by $\rho = \frac{(\text{horizontal velocity})^2}{g}$. Using relative velocity:
$$\frac{(u_x + v)^2}{g} \ge r$$ $$\frac{(R/T + v)^2}{g} \ge r$$Rearranging to solve for $R$:
$$(R/T + v)^2 \ge gr \implies R/T + v \ge \sqrt{gr}$$ $$R/T \ge \sqrt{gr} – v \implies R \ge T(\sqrt{gr} – v)$$Substitute $T = 4\sqrt{r/g}$:
$$R \ge 4\sqrt{\frac{r}{g}} (\sqrt{gr} – v)$$ $$R \ge 4 (\sqrt{\frac{r}{g}}\sqrt{gr} – v\sqrt{\frac{r}{g}})$$ $$R \ge 4 (r – v\sqrt{\frac{r}{g}}) \quad \text{or} \quad R \ge 4(\sqrt{gr} – v)\sqrt{\frac{r}{g}}$$This matches option (d).