A projectile is fired with initial speed \( 500 \mathrm{~m} / \mathrm{s} \) at an angle 35 degrees above the horizontal over flat ground. What are its maximum height and range?

We begin by resolving the initial velocity \( v_0 = 500\,\mathrm{m/s} \) into its horizontal and vertical components. The launch angle is \( \theta = 35^\circ \). 1. The vertical component is \[ v_{0y} = v_0 \sin\theta = 500 \sin 35^\circ, \] and the horizontal component is \[ v_{0x} = v_0 \cos\theta = 500 \cos 35^\circ. \] 2. The maximum height \( H \) is reached when the vertical velocity becomes zero. Using the formula \[ H = \frac{v_{0y}^2}{2g}, \] where the acceleration due to gravity is \( g = 9.8\,\mathrm{m/s^2} \), we substitute: \[ H = \frac{(500 \sin 35^\circ)^2}{2 \times 9.8}. \] Simplify the expression: \[ H = \frac{500^2 \sin^2 35^\circ}{19.6}. \] Evaluating numerically: - First, compute \( 500^2 = 250\,000 \). - Using \( \sin 35^\circ \approx 0.5736 \), we get \[ \sin^2 35^\circ \approx 0.5736^2 \approx 0.329. \] - Thus, \[ H \approx \frac{250\,000 \times 0.329}{19.6} \approx \frac{82\,250}{19.6} \approx 4194\,\mathrm{m}. \] 3. The range \( R \) of the projectile is given by \[ R = \frac{v_0^2 \sin(2\theta)}{g}. \] Substitute the known values: \[ R = \frac{250\,000 \sin 70^\circ}{9.8}. \] With \( \sin 70^\circ \approx 0.9397 \): \[ R \approx \frac{250\,000 \times 0.9397}{9.8} \approx \frac{234\,925}{9.8} \approx 23\,930\,\mathrm{m}. \] Thus, the maximum height is approximately \( 4194\,\mathrm{m} \) and the range is approximately \( 23\,930\,\mathrm{m} \).