Projectile Motion with Air Resistance

The horizontal motion has an initial velocity (v_{0x} = v_0 \cos 30^\circ = 20 \times 0.866 = 17.32, \text{m/s}). The horizontal displacement is (x = 32.0, \text{m}) and the time of flight is (t = 1.8, \text{s}). Using (x = v_{0x} t + \frac{1}{2} a_x t^2), we solve for (a_x): [32.0 = 17.32 \times 1.8 + \frac{1}{2} a_x (1.8)^2] [32.0 = 31.176 + 1.62 a_x] [0.824 = 1.62 a_x \implies a_x \approx 0.509, \text{m/s}^2] The horizontal acceleration is negative (opposite to motion). The final horizontal velocity is (v_x = v_{0x} + a_x t = 17.32 - 0.509 \times 1.8 = 17.32 - 0.916 = 16.40, \text{m/s}). Thus, the horizontal component of velocity just before landing is approximately (16.4, \text{m/s}).

The vertical motion is affected by both gravity and air resistance. The initial vertical velocity is (v_{0y} = v_0 \sin 30^\circ = 20 \times 0.5 = 10.0, \text{m/s}). The vertical displacement is zero (ground to ground). Using (y = v_{0y} t + \frac{1}{2} a_y t^2) with (y=0): [0 = 10.0 \times 1.8 + \frac{1}{2} a_y (1.8)^2] [0 = 18.0 + 1.62 a_y \implies a_y = -\frac{18.0}{1.62} \approx -11.11, \text{m/s}^2] This acceleration is the net vertical acceleration, which includes gravity ((-9.8, \text{m/s}^2)) and the vertical component of air resistance. The final vertical velocity is (v_y = v_{0y} + a_y t = 10.0 - 11.11 \times 1.8 = 10.0 - 20.0 = -10.0, \text{m/s}). In the absence of air resistance, the final vertical velocity would be (-10.0, \text{m/s}) (same magnitude, opposite direction). So, the vertical component just before landing is (-10.0, \text{m/s}), which is the same as in the vacuum case. This indicates that the net vertical acceleration is exactly (-g) on average, meaning the vertical component of air resistance is negligible or averaged to zero over the flight. However, the horizontal air resistance reduces the horizontal range.