A cannon fires a cannonball (from ground level) with an inital velocity of 300 feet per sec with an angle of inclination of 50°.

Question

A cannon fires a cannonball (from ground level) with an inital velocity of 300 feet per sec with an angle of inclination of 50°.

1. Find the parametric equations that describe the position of the object is a function of time.

2. How long is the object in the air?

3. When is the object at its maximum height? Determine the maximum height of the object.

4. Determine the distance that the object traveled.

Answer 1

1) This is standard projectile motion, except using feet instead of meters as in physics class. Given vertical acceleration a_y = -32 ft/s², v_y = ∫a_y dt = -32 t + v_o = -32 t + (300 sin 50°), Δy = ∫v_y dt = -16 t² + v_o t + y_o = -16 t² + (300 sin 50°) t + 0. Air resistance is neglected, so a_x = 0, v_x = v_o = (300 cos 50°), Δx = v_o t + x_o = (300 cos 50°) t + 0.

2) Δy = 0 = -16 t² + 229.8 t = t(-16 t +229.8); solution t = 0 is the launch, so we want -16 t +229.8 = 0, t = 14.36 s for the total duration.

3) The max height is when v_y = 0 = -32 t + 229.8, t = 7.18 s (half of the total). At this point Δy = -16 (7.18)² + 229.8 t = -825 + 1650 = 825 ft.

4) Δx = (192.8)(14.36) = 2770 ft.