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UPSC 2022 Maths Optional Paper 1 Q5d — Step-by-Step Solution

10 marks · Section B

Projectile motion · Dynamics & Statics · asked 6× in 14 yrs · Read the full method →

Question

A projectile is fired from a point $O$ with velocity $\sqrt{2gh}$ and hits a tangent at the point $P(x,y)$ in the plane, the axes $OX$ horizontal and $OY$ vertically downward. Show that if the two possible directions of projection are at right angles, then $x^2=2hy$, and one of the directions of projection bisects the angle $POX$.

Technique

Standard projectile parametric form with downward $OY$; quadratic in $\tan\alpha$; perpendicularity ⇔ product of roots = $-1$; half-angle identity ($\tan(\beta/2)=(1-\cos\beta)/\sin\beta$) to identify the bisector.

Solution

Setup. Note the axis convention: $OX$ horizontal, $OY$ vertically downward. So a projectile launched with horizontal component $u\cos\alpha$ (right) and vertical component $u\sin\alpha$ (downward, positive in this frame) is additionally accelerated downward by gravity.

Position at time $t$ (with $OY$ downward, gravity $+g$ in $y$-direction):

• $x=u\cos\alpha\cdot t$
• $y=u\sin\alpha\cdot t+\tfrac{1}{2}g t^2$

With $u^2=2gh$:

Eliminate $t=x/(u\cos\alpha)$:

Use $\sec^2\alpha=1+\tan^2\alpha$:

With $u^2=2gh$: $\dfrac{g x^2}{2u^2}=\dfrac{g x^2}{4gh}=\dfrac{x^2}{4h}$.

Step 1 — Quadratic in $\tan\alpha$

Rearrange:

Multiply by $4h/x^2$:

Two roots $\tan\alpha_1,\tan\alpha_2$.

Step 2 — Vieta

$\tan\alpha_1+\tan\alpha_2=-\dfrac{4h}{x}$.

$\tan\alpha_1\tan\alpha_2=1-\dfrac{4hy}{x^2}$.

Step 3 — Perpendicularity condition

Directions are perpendicular ⇔ $\tan\alpha_1\tan\alpha_2=-1$.

Setting $1-\dfrac{4hy}{x^2}=-1$:

Answer