CHALLENGE · VECTORS AND PROJECTILE MOTION
COMPLEMENTARY ANGLES SAME RANGE
A cannon fires at 50 m/s. The target is 200 m away on flat ground. There are two launch angles that reach exactly that range: a shallow trajectory and a steep one. Find both angles. Then compute the time of flight for each and the difference in hang time between them.
Linked equations:R = \frac{v_0^2\sin 2\theta}{g}T = \frac{2v_0\sin\theta}{g}\sin 2\theta = \sin(\pi - 2\theta)
§ 01
Step-by-step solution
Work through one named subgoal at a time. Each step is checked deterministically against the canonical solver — no AI required to verify correctness. Get an AI explanation when you're stuck.
Step 1
What value of sin(2θ) is needed to reach the target at 200 m?
Hint
Rearrange the range formula R = v₀²·sin(2θ)/g to get sin(2θ) = R·g/v₀².
Step 2
Find the shallow (low) launch angle.
Step 3
Find the steep (high) launch angle.
Step 4
What is the difference in time of flight between the high arc and the low arc?
Solution walkthrough
The range formula is R = v₀²·sin(2θ)/g. For R = 200 m, v₀ = 50 m/s, g ≈ 9.807 m/s²: sin(2θ) = 200·9.807/2500 ≈ 0.7846. So 2θ = asin(0.7846) ≈ 51.7°, giving θ_low ≈ 25.8°... let me recheck: 200×9.807 = 1961.4, 1961.4/2500 = 0.7846. asin(0.7846) ≈ 51.7°, θ_low ≈ 25.85° ≈ 0.451 rad. θ_high = 90° - 25.85° = 64.15° ≈ 1.120 rad.
Both angles share the same sin(2θ) because sin(2θ) = sin(π - 2θ), so one angle gives 2θ and the other gives 180° - 2θ, with the two θ values summing to 90°. The ranges are identical by construction.
Time of flight: T = 2·v₀·sin(θ)/g. For the shallow angle, sin(25.85°) ≈ 0.436, so T_low = 2·50·0.436/9.807 ≈ 4.45 s. For the steep angle, sin(64.15°) ≈ 0.900, so T_high = 2·50·0.900/9.807 ≈ 9.18 s. The high-arc shell hangs in the air roughly twice as long. In real artillery this matters enormously: the low arc is fast and direct; the high arc drops nearly vertically and can clear obstacles but gives defenders more time to react.
§ 02
Try it with AI
Continue the conversation with the Physics tutor — the problem context is pre-loaded.
Open in Physics.Ask