Projectile Motion and Energy in a School Sports Launch
A physics club builds a spring-loaded launcher for a school competition. A ball of mass 0.20 kg is launched from ground level at 12 m/s at an angle of 40° above the horizontal. The launcher spring has a spring constant of 800 N/m and is compressed by 0.15 m before release. Air resistance is neglected. The ball must clear a 2.5 m high wall located 10 m horizontally from the launch point. Take g = 9.8 m/s². sin 40° ≈ 0.643, cos 40° ≈ 0.766.
1. Determine whether the ball clears the wall. Show the trajectory calculation and state the margin.
Read the worked answer
Velocity components: v₀ₓ = 12 × cos 40° = 12 × 0.766 = 9.19 m/s v₀ᵧ = 12 × sin 40° = 12 × 0.643 = 7.72 m/s
Time to reach x = 10 m: t = x / v₀ₓ = 10 / 9.19 = 1.088 s
Height at that time: y = v₀ᵧ t − ½ g t² y = 7.72 × 1.088 − 0.5 × 9.8 × (1.088)² y = 8.40 − 0.5 × 9.8 × 1.184 y = 8.40 − 5.80 = 2.60 m
Conclusion: The ball reaches 2.60 m at the wall, clearing the 2.5 m wall by about 0.10 m. The margin is small (~10 cm), so any air resistance or launch-angle error could cause failure.
2. Compare the elastic potential energy stored in the spring with the kinetic energy of the ball at launch, and explain any discrepancy in terms of a realistic launcher.
Read the worked answer
Elastic potential energy: Eₚ = ½ k x² = 0.5 × 800 × (0.15)² = 0.5 × 800 × 0.0225 = 9.0 J
Kinetic energy at launch: Eₖ = ½ m v² = 0.5 × 0.20 × (12)² = 0.5 × 0.20 × 144 = 14.4 J
Discrepancy: The kinetic energy (14.4 J) exceeds the spring's stored elastic energy (9.0 J). In an ideal launcher, energy conservation would require Eₖ ≤ Eₚ. The excess of 5.4 J means the model is inconsistent: either the spring constant, compression, or launch speed is misreported, or another energy source exists (e.g., the spring is not the only source - a compressed-air assist, or the ball is pushed by a moving piston with additional stored energy).
A realistic launcher also loses energy to friction and to the mass of the spring/piston itself, so Eₖ should be less than Eₚ, not more. The most likely explanation is that the quoted 12 m/s is too high for the stated spring, or the spring constant is understated. This is a useful sanity check: energy conservation bounds the maximum possible launch speed to v = √(2Eₚ/m) = √(2×9.0/0.20) = √90 ≈ 9.5 m/s.