Question

A rock is thrown at a window that is located 18m above the ground. The rock is thrown at an angle 40 degrees above the horizontal. The rock is thrown from a height of 2.00m above the ground with a speed of 30m/s and experiences no appreciable air resistance. If the rock strikes the window on its upward trajectory, from what horizontal distance from the window was it released???

Answer 1

Answer on Question#49708 – Physics – Mechanics | Kinematics | Dynamics

$$L = 30 \cos (40{}^{\circ}) \times \left(3 \sin (40{}^{\circ}) - \sqrt{9 \sin^{2} (40{}^{\circ}) - \frac{16}{5}}\right) \approx 27.77$$

Solution

Equations of motion:

$$m \frac{d^{2} x}{d t^{2}} = 0$$ $$m \frac{d^{2} y}{d t^{2}} = - m g$$

Initial conditions:

$$x = 0; \quad V_{x}(0) = 30 \cos (40{}^{\circ}); \quad y(0) = 2; \quad V_{y}(0) = 30 \sin (40{}^{\circ});$$

Explicit solutions of the equations above:

$$x = C_{1} + C_{2} t$$ $$y = - \frac{g t^{2}}{2} + C_{3} t + C_{4}$$

Use initial conditions:

$$x(0) = C_1 + C_2(0) = C_1 = 0$$ $$V_x(0) = \frac{dx}{dt}(0) = C_2 = 30 \cos(40{}^\circ)$$ $$y(0) = -\frac{g(0)^2}{2} + C_3(0) + C_4 = C_4 = 2$$ $$V_y(0) = \frac{dy}{dt}(0) = -g(0) + C_3 = C_3 = 30 \sin(40{}^\circ)$$

Thus, we have:

$$x(t) = 30 \cos(40{}^\circ) t$$ $$y(t) = 2 + 30 \sin(40{}^\circ) t - \frac{g t^2}{2}$$

Assume $g = 10$

$$y(t) = 2 + 30 \sin(40{}^\circ) t - 5 t^2$$

Our aim to define such distance $L$ that at some time $t_1$: $y(t_1) = 18$; $V_y(t_1) \geq 0$.

$$2 + 30 \sin(40{}^\circ) t_1 - 5 t_1^2 = 18$$ $$-5 t_1^2 + 30 \sin(40{}^\circ) t_1 - 16 = 0$$ $$t_1^2 - 6 \sin(40{}^\circ) t_1 + \frac{16}{5} = 0$$ $$D = 36 \sin^2(40{}^\circ) - \frac{64}{5}$$ $$t_1(\downarrow) = \frac{6 \sin(40{}^\circ) + \sqrt{36 \sin^2(40{}^\circ) - \frac{64}{5}}}{2} = 3 \sin(40{}^\circ) + \sqrt{9 \sin^2(40{}^\circ) - \frac{16}{5}}$$ $$t_1(\uparrow) = 3 \sin(40{}^\circ) - \sqrt{9 \sin^2(40{}^\circ) - \frac{16}{5}}$$

Obviously, $t_1(\uparrow)$ refers to upward trajectory, $t_1(\downarrow)$ – downward trajectory.

Determine $L = x(t_1)$

$$L = 30 \cos(40{}^\circ) \times \left(3 \sin(40{}^\circ) - \sqrt{9 \sin^2(40{}^\circ) - \frac{16}{5}}\right) \approx 27.77$$

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