Question pattern: Projectile launched above horizontal
Original question 1 of 2 in this pattern
Angled projectile returning to launch height
A projectile is launched at 20.0 m s⁻¹ at an angle of 35.0° above the horizontal and lands at the same height. Air resistance is negligible. Take g = 9.81 m s⁻².
Calculate:
a) the initial horizontal and vertical velocity components;
b) the maximum height above the launch point;
c) the total time of flight;
d) the horizontal range.
Worked answer
Solution
- Step 1
Resolve the launch velocity:
uₓ = 20.0 cos 35.0° = 16.4 m s⁻¹;uᵧ = 20.0 sin 35.0° = 11.5 m s⁻¹. - Step 2
At maximum height,
vᵧ = 0. Usevᵧ² = uᵧ² + 2aᵧsᵧ:h = uᵧ² / (2g) = 6.71 m. - Step 3
Time to maximum height:
t_up = uᵧ / g = 1.17 s. - Step 4
The landing height equals the launch height, so total flight time is
2t_up = 2.34 s. - Step 5
Horizontal range:
R = uₓt = 16.4 × 2.34 = 38.3 musing unrounded values.
Final answer
uₓ = 16.4 m s⁻¹; uᵧ = 11.5 m s⁻¹; maximum height = 6.71 m; flight time = 2.34 s; range = 38.3 m.
Common mistake
Using the full 20.0 m s⁻¹ as the vertical speed when finding the maximum height.
Concepts tested
- projectile motion
- velocity components
- maximum height
- time of flight
- range
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