Projectile hitting target 600 m below

Setting up

Origin at the point of projection, v₀ = 61 m/s, angle θ above the horizontal. The target lies 600 m below the horizontal axis on an incline. Trajectory:

y = x tan θ − g x²/(2v₀² cos²θ) = x tan θ − [g x²/(2v₀²)](1 + tan²θ)

Position of the target

The slope is quoted as 3/4.

• If the incline falls 3 m for every 4 m horizontally, the target is at x = 800 m. The least speed that can reach (800, −600) is √[g(√(800² + 600²) − 600)] = 62.6 m/s, which is more than 61 m/s. The discriminant is negative and no firing angle exists, so this reading cannot be the one intended.
• With the slope read as 3 horizontal to 4 vertical (the incline falls 4 m for every 3 m of horizontal distance), the target is at x = 600 × 3/4 = 450 m and the problem has a solution. This is used below.

Firing angle

g x²/(2v₀²) = 9.81 × 450²/(2 × 61²) = 266.93

−600 = 450 tan θ − 266.93(1 + tan²θ) 266.93 tan²θ − 450 tan θ − 333.07 = 0 tan θ = [450 ± √(450² + 4 × 266.93 × 333.07)]/(2 × 266.93) = (450 ± 747.1)/533.87

tan θ = 2.2423 or −0.5565, so θ = 65.96° above the horizontal (the second root, θ = −29.1°, means firing below the horizontal).

Maximum height from the target

• Vertical component of velocity: v_y = 61 sin 65.96° = 55.7 m/s
• Rise above the axis: h = v_y²/(2g) = 55.7²/19.62 = 158.2 m
• Height above the target = 158.2 + 600 = 758.2 m

Answer: θ ≈ 66.0°; maximum height above the target ≈ 758 m. (Time of flight = 18.1 s and 61 cos 65.96° × 18.1 = 450 m, which checks the target distance.)