91Ó°ÊÓ

We need to find the horizontal component (v_x) and the vertical component (v_y) of the initial velocity. We can do this by using the given initial speed (v) and angle (θ) above the horizontal, and applying trigonometric functions. \(v_x = v \cdot \cos{\theta}\) \(v_y = v \cdot \sin{\theta}\) Plugging in the values, we get: \(v_x = 50.0 \mathrm{m/s} \cdot \cos{30^\circ} \approx 43.30 \mathrm{m/s}\) \(v_y = 50.0 \mathrm{m/s} \cdot \sin{30^\circ} = 25.0 \mathrm{m/s}\)

By knowing the horizontal range (R) and horizontal speed (v_x), we can find the time of flight (t) as: \(t = \dfrac{R}{v_x}\) \(t = \dfrac{2165 \mathrm{m}}{43.30 \mathrm{m/s}} \approx 50.0 \mathrm{s}\)

Now we'll use the vertical motion equation to determine the gravitational acceleration (g) of the planet: \(v_y^2 = u_y^2 - 2g \cdot h\) Since the projectile motion starts and ends at the same height (flat ground), the vertical displacement (h) is 0. The formula now becomes: \(v_y^2 = u_y^2 - 2g \cdot 0\) \(v_y^2 = u_y^2\) Since both \(v_y\) (vertical velocity at the highest point) and \(u_y\) (initial vertical velocity) are equal, we can use another vertical motion formula: \(h = u_yt - \dfrac{1}{2}gt^2\) Once again, since h = 0, the equation simplifies to: \(0 = u_yt - \dfrac{1}{2}gt^2\) Solving for g, we get: \(g = \dfrac{2u_yt}{t^2}\) Plugging in the values for \(u_y\) (initial vertical velocity) and \(t\) (time of flight), we get: \(g = \dfrac{2 \cdot 25.0 \mathrm{m/s} \cdot 50.0 \mathrm{s}}{(50.0 \mathrm{s})^2} = \dfrac{2500 \mathrm{m}}{2500 \mathrm{s^2}}\) \(g = 1.0 \mathrm{m/s^2}\) The gravitational acceleration of the newly discovered planet is 1.0 m/s².