Physics Projectile Motion

Question

An aircraft flies at a height of 2000m with a constant velocity of 150 mps in a straight horizontal line. As it passes vertically over a cannon, a shell is fired. Find the minimum muzzle speed and the angle from the horizontal to hit the plane.

Answer 1

Interesting problem! So v_x = 150 m/s, and Δy for the projectile is +2000 m. Use the equation v_f² = v_y² + 2 a Δy, where v_y is the initial vertical component of the muzzle speed v_o, and acceleration is taken to be -9.8 m/s². Let final vertical speed be zero to achieve the minimum angle and speed, so that the projectile just gets to the height of the plane: 0 = v_y² - 2*9.8*2000 and v_y² = 39,200. In these problems there is no horizontal acceleration, so this component of v_o just needs to match that of the plane. The muzzle speed then comes from v_o² = v_x² + v_y² = 22,500 + 39,200 --> v_o = 248 m/s.

To get the angle, sketch the horizontal and vertical components of v_o to see that tan θ = v_y / v_x = 198 / 150 = 1.320, and the angle is θ = tan^-1 (1.320) = 52.9°.