Key Concepts and Equations The motion of a projectile launched horizontally from height $H$ with velocity $v$ is described by:

• Horizontal position: $x(t) = vt$
• Vertical position: $y(t) = H - \frac{1}{2}gt^2$ Eliminating time $t=x/v$ gives the trajectory equation:

$$y(x) = H - \frac{g}{2v^2}x^2$$

A perfect elastic collision with the horizontal ground reverses the vertical component of velocity while keeping the horizontal component constant.

[Q1] Find the ratio $v_1/v_2$. Both balls are launched from the same height $H$. The time of flight $T$ to the ground is determined solely by the vertical motion: $H = \frac{1}{2}gT^2$, so $T = \sqrt{2H/g}$ for both balls' initial descent.

For Ball 1, it lands at a horizontal distance $R$.

$$R = v_1 T = v_1 \sqrt{\frac{2H}{g}} \quad (1)$$

For Ball 2, it first lands at a horizontal distance $R/3$.

$$\frac{R}{3} = v_2 T = v_2 \sqrt{\frac{2H}{g}} \quad (2)$$

To find the ratio $v_1/v_2$, we divide equation (1) by equation (2):

$$\frac{R}{R/3} = \frac{v_1 \sqrt{2H/g}}{v_2 \sqrt{2H/g}}$$ $$3 = \frac{v_1}{v_2}$$

[Q2] Find the position of the rod, $x_P$. First, we establish the trajectory equations for both paths that clear the rod.

Trajectory of Ball 1: From equation (1), we can express the term $\frac{g}{2v_1^2}$ as $\frac{H}{R^2}$. Substituting this into the general trajectory equation gives:

$$y_1(x) = H - H\frac{x^2}{R^2} = H\left(1 - \frac{x^2}{R^2}\right) \quad (3)$$

Trajectory of Ball 2 (rebound): Ball 2's rebound starts at $(R/3, 0)$ and ends at $(R, 0)$. Due to the perfect elastic collision, the rebound trajectory is a symmetric parabola. Its vertex (maximum height) occurs at the midpoint of its horizontal range.

• Vertex x-coordinate: $x_v = \frac{R/3 + R}{2} = \frac{2R}{3}$
• The peak height of the rebound is equal to the initial drop height $H$ (by conservation of energy, or time-reversal symmetry). So, the vertex is at $(2R/3, H)$. The equation for a parabola with vertex $(x_v, y_v)$ is $y = a(x-x_v)^2 + y_v$.

$$y_{reb}(x) = a\left(x - \frac{2R}{3}\right)^2 + H$$

Since the parabola passes through $(R, 0)$:

$0 = a\left(R - \frac{2R}{3}\right)^2 + H \implies 0 = a\left(\frac{R}{3}\right)^2 + H \implies a = -\frac{9H}{R^2}$

So, the rebound trajectory is:

$$y_{reb}(x) = H - \frac{9H}{R^2}\left(x - \frac{2R}{3}\right)^2 \quad (4)$$

Both trajectories pass through the point $(x_P, h)$. Therefore, $h = y_1(x_P) = y_{reb}(x_P)$.

$$H\left(1 - \frac{x_P^2}{R^2}\right) = H - \frac{9H}{R^2}\left(x_P - \frac{2R}{3}\right)^2$$

Canceling $H$ and simplifying:

$$-\frac{x_P^2}{R^2} = -\frac{9}{R^2}\left(x_P^2 - \frac{4R}{3}x_P + \frac{4R^2}{9}\right)$$ $$x_P^2 = 9\left(x_P^2 - \frac{4R}{3}x_P + \frac{4R^2}{9}\right)$$ $$x_P^2 = 9x_P^2 - 12Rx_P + 4R^2$$ $$8x_P^2 - 12Rx_P + 4R^2 = 0$$

Dividing by 4:

$$2x_P^2 - 3Rx_P + R^2 = 0$$

Factoring the quadratic equation:

$$(2x_P - R)(x_P - R) = 0$$

The solutions are $x_P = R$ or $x_P = R/2$. From the figure, the rod is located before the final landing point, so $x_P eq R$.

$$x_P = \frac{R}{2}$$

[Q3] Find the height of the rod, $h$. Substitute the value of $x_P = R/2$ into the simpler trajectory equation for Ball 1 (equation 3):

$$h = y_1(x_P) = H\left(1 - \frac{x_P^2}{R^2}\right)$$ $$h = H\left(1 - \frac{(R/2)^2}{R^2}\right) = H\left(1 - \frac{R^2/4}{R^2}\right)$$ $$h = H\left(1 - \frac{1}{4}\right) = \frac{3}{4}H$$