Question

Two points A and B are located on the ground a certain distance $10\sqrt{2}$ m apart. Two rocks are launched simultaneously from points A and B with equal speeds but at different angles. Each rocks lands at the launch point of the other. Knowing that one of the rocks is launched at an angle $\theta = 37^\circ$ with the horizontal, what is the minimum distance between the rocks during the flight? (given: cos37° = 4/5)

• A. 2 m
• B. 4 m
• C. 6 m
• D. 8 m

Answer 1

Answer: (A)

2 m

Answer 2

The problem describes a scenario of projectile motion where two rocks are launched simultaneously from points A and B, with a distance $R = 10\sqrt{2}$ m between them. Both rocks are launched with the same initial speed $v_0$ but at different angles. Crucially, each rock lands at the launch point of the other. This implies that the horizontal range of both projectiles is equal to the distance $R$.

Let the launch angles be $\theta_1$ and $\theta_2$. The horizontal range $R$ of a projectile launched with speed $v_0$ at an angle $\theta$ is given by the formula: $R = \frac{v_0^2 \sin(2\theta)}{g}$

Since both rocks have the same range $R$ and the same initial speed $v_0$, we have: $\frac{v_0^2 \sin(2\theta_1)}{g} = \frac{v_0^2 \sin(2\theta_2)}{g}$ This simplifies to $\sin(2\theta_1) = \sin(2\theta_2)$. Given that the angles are different, the relationship between them must be $2\theta_1 = 180^\circ - 2\theta_2$, which leads to $\theta_1 + \theta_2 = 90^\circ$.

We are given that one of the rocks is launched at an angle $\theta_1 = 37^\circ$. Therefore, the other launch angle is $\theta_2 = 90^\circ - 37^\circ = 53^\circ$.

We are also given $\cos(37^\circ) = 4/5$. From this, we can deduce $\sin(37^\circ) = \sqrt{1 - (\cos(37^\circ))^2} = \sqrt{1 - (4/5)^2} = \sqrt{1 - 16/25} = \sqrt{9/25} = 3/5$. Using the complementary angle identities, $\sin(53^\circ) = \cos(37^\circ) = 4/5$ and $\cos(53^\circ) = \sin(37^\circ) = 3/5$.

Now, let's set up a coordinate system. Let point A be the origin (0, 0) and point B be at $(R, 0)$.

The position of the rock launched from A (Rock 1) at time $t$ is: $x_1(t) = v_0 \cos(\theta_1) t = v_0 \cos(37^\circ) t = \frac{4}{5} v_0 t$ $y_1(t) = v_0 \sin(\theta_1) t - \frac{1}{2}gt^2 = v_0 \sin(37^\circ) t - \frac{1}{2}gt^2 = \frac{3}{5} v_0 t - \frac{1}{2}gt^2$

The rock launched from B (Rock 2) starts at $(R, 0)$ and travels towards A. Its position at time $t$ is: $x_2(t) = R - v_0 \cos(\theta_2) t = R - v_0 \cos(53^\circ) t = R - \frac{3}{5} v_0 t$ $y_2(t) = v_0 \sin(\theta_2) t - \frac{1}{2}gt^2 = v_0 \sin(53^\circ) t - \frac{1}{2}gt^2 = \frac{4}{5} v_0 t - \frac{1}{2}gt^2$

The time of flight for Rock 1 is $T_1 = \frac{2v_0 \sin(\theta_1)}{g} = \frac{2v_0 (3/5)}{g} = \frac{6v_0}{5g}$. The time of flight for Rock 2 is $T_2 = \frac{2v_0 \sin(\theta_2)}{g} = \frac{2v_0 (4/5)}{g} = \frac{8v_0}{5g}$. Since $T_1 < T_2$, both rocks are in flight during the interval $[0, T_1]$.

We are interested in the distance between the rocks. Let's find the difference in their coordinates: $\Delta x(t) = x_1(t) - x_2(t) = \frac{4}{5} v_0 t - (R - \frac{3}{5} v_0 t) = (\frac{4}{5} + \frac{3}{5}) v_0 t - R = \frac{7}{5} v_0 t - R$ $\Delta y(t) = y_1(t) - y_2(t) = (\frac{3}{5} v_0 t - \frac{1}{2}gt^2) - (\frac{4}{5} v_0 t - \frac{1}{2}gt^2) = (\frac{3}{5} - \frac{4}{5}) v_0 t = -\frac{1}{5} v_0 t$

The square of the distance between the rocks is $d(t)^2 = (\Delta x(t))^2 + (\Delta y(t))^2$: $d(t)^2 = (\frac{7}{5} v_0 t - R)^2 + (-\frac{1}{5} v_0 t)^2$ $d(t)^2 = (\frac{49}{25} v_0^2 t^2 - 2 \cdot \frac{7}{5} v_0 t R + R^2) + \frac{1}{25} v_0^2 t^2$ $d(t)^2 = \frac{50}{25} v_0^2 t^2 - \frac{14}{5} v_0 R t + R^2$ $d(t)^2 = 2 v_0^2 t^2 - \frac{14}{5} v_0 R t + R^2$

This is a quadratic equation in $t$ of the form $at^2 + bt + c$, where $a = 2v_0^2$, $b = -\frac{14}{5} v_0 R$, and $c = R^2$. The minimum value of this quadratic occurs at $t_{min} = -b/(2a)$. $t_{min} = -(-\frac{14}{5} v_0 R) / (2 \cdot 2 v_0^2) = \frac{14 v_0 R}{5 \cdot 4 v_0^2} = \frac{14 R}{20 v_0} = \frac{7R}{10 v_0}$

To ensure this minimum occurs during the flight, we compare $t_{min}$ with $T_1$. We need to find $v_0$. The range $R$ is given by $R = \frac{v_0^2 \sin(2\theta_1)}{g}$. $10\sqrt{2} = \frac{v_0^2 \sin(2 \times 37^\circ)}{g} = \frac{v_0^2 \sin(74^\circ)}{g}$ $\sin(74^\circ) = 2 \sin(37^\circ) \cos(37^\circ) = 2 \times (3/5) \times (4/5) = 24/25$. So, $10\sqrt{2} = \frac{v_0^2 (24/25)}{g}$, which means $\frac{v_0^2}{g} = \frac{10\sqrt{2} \times 25}{24} = \frac{250\sqrt{2}}{24} = \frac{125\sqrt{2}}{12}$.

Now, let's express $T_1$ in terms of $R$ and $v_0$: $T_1 = \frac{6v_0}{5g}$. We can write $g = \frac{v_0^2 (24/25)}{R}$. $T_1 = \frac{6v_0}{5 \left(\frac{v_0^2 (24/25)}{R}\right)} = \frac{6v_0 R}{5 v_0^2 (24/25)} = \frac{6R}{5 v_0 (24/25)} = \frac{6R}{v_0 (120/25)} = \frac{6R}{v_0 (24/5)} = \frac{30R}{24v_0} = \frac{5R}{4v_0}$.

Comparing $t_{min}$ and $T_1$: $t_{min} = \frac{7R}{10v_0}$ $T_1 = \frac{5R}{4v_0} = \frac{12.5R}{10v_0}$ Since $\frac{7R}{10v_0} < \frac{12.5R}{10v_0}$, the minimum distance occurs within the flight time of both rocks.

Now, substitute $t_{min}$ back into the equation for $d(t)^2$: $d_{min}^2 = 2 v_0^2 (\frac{7R}{10 v_0})^2 - \frac{14}{5} v_0 R (\frac{7R}{10 v_0}) + R^2$ $d_{min}^2 = 2 v_0^2 \frac{49R^2}{100 v_0^2} - \frac{98 R^2}{50} + R^2$ $d_{min}^2 = \frac{98 R^2}{100} - \frac{98 R^2}{50} + R^2$ $d_{min}^2 = \frac{49 R^2}{50} - \frac{98 R^2}{50} + \frac{50 R^2}{50}$ $d_{min}^2 = \frac{(49 - 98 + 50) R^2}{50} = \frac{1 R^2}{50}$

The minimum distance $d_{min}$ is the square root of $d_{min}^2$: $d_{min} = \sqrt{\frac{R^2}{50}} = \frac{R}{\sqrt{50}} = \frac{R}{5\sqrt{2}}$

Finally, substitute the given value of $R = 10\sqrt{2}$ m: $d_{min} = \frac{10\sqrt{2} \text{ m}}{5\sqrt{2}} = 2 \text{ m}$

The minimum distance between the rocks during the flight is 2 m.