A Physics student had just learnt about the projectile motion in an ideal environment (no air resistance). After calculating the expected results, the student heads outdoors to test the theory in an open field where a strong head wind was blowing horizontally against the projectile's motion. Which one of the following observations is to be expected?

Question

BlackTom AI · Accepted Answer

Horizontal component of the projectile’s velocity is reduced by a headwind, so the horizontal distance (range) will be smaller than the no‑air (ideal) prediction.

1) Relevant concepts
- Decompose initial speed u into horizontal u_x = u cosθ and vertical u_y = u sinθ.
- In ideal projectile motion (no air resistance) horizontal velocity is constant and range R_ideal = u_x · time_of_flight. With gravity g, time_of_flight = 2u_y/g, so R_ideal = u_x·(2u_y/g) = (u^2 sin2θ)/g.
- A headwind produces a horizontal opposing velocity/drag, reducing net horizontal speed and so reducing range. Vertical motion is controlled by gravity; a purely horizontal wind does not change the gravity term, though real air resistance can slightly change vertical motion.

2) Step‑by‑step reasoning
- Start with components: u_x = u cosθ, u_y = u sinθ.
- Ideal range: R_ideal = u_x · (2u_y/g) = (u^2 sin2θ)/g.
- Introduce a steady headwind of speed w directed opposite motion. The projectile’s ground‑relative horizontal speed becomes approximately (u_x − w) (for moderate w; strong wind complicates with drag but sign is the same).
- Time of flight (to first approximation, neglecting vertical drag) remains t = 2u_y/g because vertical motion is governed by gravity.

3) Intermediate calculation
R_actual ≈ (u_x − w) · t = (u cosθ − w) · (2u sinθ / g)
= (u^2 sin2θ / g) − (2w u sinθ / g).
The second term is positive for w>0, so R_actual < R_ideal.

4) Why the correct answer is correct
Because the headwind subtracts from the horizontal component of velocity (or produces a horizontal decelerating force), the projectile travels more slowly forward and lands sooner, reducing the horizontal range relative to the ideal no‑air prediction.

5) Why other choices are incorrect
- (a) More range is wrong because an opposing wind cannot increase forward horizontal speed.
- (b) Maximum height greater is unlikely; a horizontal headwind does not add upward velocity.
- (c) Maximum height less is possible if there is significant vertical air resistance, but the primary and most direct, guaranteed effect of a horizontal headwind is reduction of range; the question emphasizes horizontal wind and the expected observation is reduced range.

Conclusion: The actual range will be less than the ideal calculated range, confirming answer d.

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