Question pattern: Projectile launched above horizontal
Original question 2 of 2 in this pattern
Angled projectile from a platform
A projectile is launched from a platform 6.0 m above level ground at 18.0 m s⁻¹ and 40.0° above the horizontal. Air resistance is negligible. Take g = 9.81 m s⁻².
Calculate:
a) the time taken to reach the ground;
b) the horizontal distance travelled;
c) the speed and direction of the velocity immediately before impact.
Worked answer
Solution
- Step 1
Resolve the launch velocity:
uₓ = 18.0 cos 40.0° = 13.79 m s⁻¹;uᵧ = 18.0 sin 40.0° = 11.57 m s⁻¹. - Step 2
Vertical motion to the ground satisfies
-6.0 = 11.57t - 4.905t². - Step 3
Solving the quadratic gives the physical root
t = 2.80 s. - Step 4
Horizontal distance:
x = uₓt = 13.79 × 2.80 = 38.6 musing unrounded values. - Step 5
Vertical impact velocity:
vᵧ = uᵧ - gt = 11.57 - 9.81(2.80) = -15.86 m s⁻¹. - Step 6
Impact speed:
v = √(13.79² + 15.86²) = 21.0 m s⁻¹. - Step 7
Direction below horizontal:
θ = tan⁻¹(15.86 / 13.79) = 49.0°.
Final answer
Time = 2.80 s; horizontal distance = 38.6 m; impact velocity = 21.0 m s⁻¹ at 49.0° below the horizontal.
Common mistake
Assuming that the flight is symmetric even though the landing point is lower than the launch point.
Concepts tested
- projectile motion
- unequal heights
- quadratic time
- impact velocity
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