Projectile Motion: Conceptual Doubts Cleared (6)

hard • 3 min read
Tags Projectile Motion

Question

A projectile is launched with speed uu at angle θ\theta above the horizontal from the ground. (a) Show that the trajectory is a parabola. (b) Prove that the range is maximum at θ=45∘\theta = 45^\circ. (c) For the same uu, do θ=30∘\theta = 30^\circ and θ=60∘\theta = 60^\circ give the same range? Explain.

Solution — Step by Step

Take launch point as origin, horizontal as xx, vertical as yy, with gg downward.

x(t)=ucos⁡θ⋅tx(t) = u\cos\theta \cdot t, y(t)=usin⁡θ⋅t−12gt2y(t) = u\sin\theta \cdot t - \tfrac{1}{2}gt^2.

From the first equation, t=x/(ucos⁡θ)t = x/(u\cos\theta). Substituting,

y=xtan⁡θ−g x22u2cos⁡2θy = x\tan\theta - \frac{g\,x^2}{2u^2\cos^2\theta}

This is a quadratic in xx, hence a parabola opening downward.

Set y=0y = 0 at landing. Solving gives

R=u2sin⁡2θgR = \frac{u^2 \sin 2\theta}{g}

Maximum of sin⁡2θ\sin 2\theta is 11, attained at 2θ=90∘2\theta = 90^\circ, i.e. θ=45∘\theta = 45^\circ. So Rmax⁡=u2/gR_{\max} = u^2/g.

sin⁡(2×30∘)=sin⁡60∘\sin(2 \times 30^\circ) = \sin 60^\circ and sin⁡(2×60∘)=sin⁡120∘=sin⁡60∘\sin(2 \times 60^\circ) = \sin 120^\circ = \sin 60^\circ. Both give the same range. Two angles complementary to each other (summing to 90∘90^\circ) always produce equal ranges.

Final answers: (a) parabola — see step 2; (b) max at 45∘45^\circ; (c) yes, 30∘30^\circ and 60∘60^\circ give equal ranges.

Why This Works

Horizontal motion has zero acceleration (we ignore air resistance), so it’s uniform. Vertical motion has constant downward gg. Two independent motions linked only by time — that decoupling is the soul of projectile motion.

The equal-range pair (θ\theta and 90∘−θ90^\circ - \theta) makes physical sense: a low, fast trajectory lands at the same spot as a high, slow one because the products of horizontal speed and time of flight come out equal.

Alternative Method

Use energy + kinematics. Time of flight T=2usin⁡θ/gT = 2u\sin\theta/g. Range R=ucos⁡θ⋅T=u2sin⁡2θ/gR = u\cos\theta \cdot T = u^2 \sin 2\theta / g. Same formula, derived without eliminating time.

The complementary-angle property is JEE Main bait. If a question asks “for what other angle is the range the same as for θ\theta?”, the answer is 90∘−θ90^\circ - \theta. No calculation needed.

Common Mistake

Using “speed at maximum height equals zero” — wrong. The vertical component is zero at the peak, but the horizontal component ucos⁡θu\cos\theta is unchanged. So the speed at the top is ucos⁡θu\cos\theta, not zero.

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