Motion in a Plane · Lesson 10 of 13
Projectile Problems with Heights and Constraints
“Solve horizontal launches, unequal landing heights and ceiling-limited range using the underlying component equations.”
• Solve a horizontal launch from a height. • Use the landing condition to find time when heights differ. • Maximise range subject to a ceiling with explicit assumptions. • Check problem geometry before inferring motion from an observed angle.
Start from the landing condition
When launch and landing heights differ, write the horizontal and vertical time equations first. Use the vertical equation to find when the projectile reaches the landing plane. Select the future root that belongs to the physical flight, then multiply by horizontal velocity to find range. The formulas T = 2u sinθ/g and R = u²sin2θ/g are consequences of equal heights; using them on a cliff launch discards the height information.
A horizontal launch
The flight-time expression comes from −h = −gT²/2 with origin at launch and upward positive. At impact v_y = −gT, while v_x = u. Squaring and adding the velocity components gives u² + g²T² = u² + 2gh. The direction below horizontal follows from tanφ = |v_y|/u. Horizontal speed changes range but not fall time for the same drop height.
Problem
A stone is launched horizontally at 15 m/s from 490 m above the ground. Take g = 9.8 m/s². Find flight time and impact speed.
- 1.T = √(2 × 490/9.8) = 10 s. Horizontal range is 15 × 10 = 150 m.
- 2.At impact, v_x = 15 m/s and v_y = −98 m/s.
- 3.Impact speed is √(15² + 98²) = √9829 = 99.14 m/s, approximately 99 m/s.
- 4.Impact direction is arctan(98/15) = 81.30° below horizontal. The horizontal component remains 15 m/s, even though the total speed becomes much larger.
A general upward launch from height
Problem
A ball is launched from 20 m above ground with velocity 12 i-hat + 15 j-hat m/s. Take g = 10 m/s². Find flight time and range.
- 1.The landing equation is 0 = 20 + 15T − 5T².
- 2.T² − 3T − 4 = 0 gives roots T = 4 s and T = −1 s. The flight after release ends at 4 s.
- 3.Range is 12 × 4 = 48 m. Impact vertical velocity is 15 − 10 × 4 = −25 m/s.
- 4.The peak is 15²/(2 × 10) = 11.25 m above launch, or 31.25 m above ground. An equal-height return would occur earlier than ground impact.
A ceiling can prevent the usual best angle
Suppose launch and landing are at floor level, speed is fixed, and a ceiling lies H_c above the launch point. A projectile clears the ceiling only if its peak height u²sin²θ/(2g) does not exceed H_c. Without this constraint, the best angle is 45°. If its peak u²/(4g) exceeds the ceiling, the allowable angle below 45° is limited by sin²θ ≤ 2gH_c/u². On the interval 0° to 45°, range increases with angle, so the largest allowed angle gives the limiting best range.
Problem
Find the largest level-floor range for launch speed 40 m/s below a ceiling 25 m above release. Take g = 9.8 m/s².
- 1.The unconstrained 45° peak would be 1600/(4 × 9.8) = 40.82 m, so that trajectory exceeds the ceiling.
- 2.The limiting upward velocity is u_y = √(2gH_c) = √490 = 22.136 m/s. Its horizontal component is √(1600 − 490) = √1110 = 33.317 m/s.
- 3.The limiting angle is arcsin(√490/40) = 33.60°. Flight time to the same floor level is 2√490/9.8 = 4.518 s.
- 4.The limiting range is 2√(490 × 1110)/9.8 = 150.51 m. If “without hitting” requires strict clearance, choose a slightly smaller angle: this value is approached as clearance tends to zero.
- 5.This result assumes no obstacles except the horizontal ceiling, the same launch and landing level, and negligible ball radius. A release point above the floor would change the available height and landing equation.
Geometry must be specified in an angular observation
An angle swept across an observer’s view does not automatically equal a distance divided by height. A straight-flying aircraft 3400 m above ground can subtend 30° over 10 s in more than one geometry. The observer’s position relative to the path and whether the interval is symmetric about overhead passage affect the horizontal distance. Stating the geometry is necessary before calculating a unique speed.
Problem
Assume an aircraft flies straight and horizontally at height 3400 m, with two positions 10 s apart symmetric about the point directly above the observer. Their sightlines enclose 30°. Find speed under these stated assumptions.
- 1.Each sightline is 15° from vertical. Each half of the horizontal travel is 3400 tan15°.
- 2.Total travel is 2 × 3400 tan15° = 1822.05 m. Speed is 1822.05/10 = 182.21 m/s, about 656 km/h.
- 3.If one observation is directly overhead and the other is 30° from vertical, the same height and elapsed time give 3400 tan30°/10 = 196.30 m/s instead.
- 4.The differing values show why the angle, height and time alone do not determine a unique speed without a geometric assumption. This is a straight-motion observation problem, not a projectile formula.
Quiz
For a horizontal launch from a fixed height, doubling the initial horizontal speed does what in the ideal model?
Key Takeaways
• Use the actual landing height to find flight time. • A horizontal launch has zero initial vertical velocity. • At unequal heights, solve the vertical quadratic and choose the physical root. • A ceiling limits the vertical launch component and can shift the best angle. • Observed angular changes require a stated geometry before speed can be inferred.