UPSC 2018 Maths Optional Paper 1 Q5e — Step-by-Step Solution

10 marks · Section B

Projectile motion · Dynamics & Statics · asked 5× in 13 yrs · Read the full method →

Question

A particle projected from a given point on the ground just clears a wall of height $h$ at a distance $d$ from the point of projection. If the particle moves in a vertical plane and if the horizontal range is $R$ , find the elevation of the projection.

Technique

Use the range-factored trajectory $y=\dfrac{\tan\alpha}{R}x(R-x)$ , then substitute the single point $(d,h)$ .

Solution

Let $\alpha$ be the angle of projection and $u$ the speed. Take the launch point as origin.

Step 1 — Trajectory equation

The path of a projectile is

y=x\tan\alpha-\frac{g\,x^2}{2u^2\cos^2\alpha}.

The horizontal range $R$ is the non-zero root of $y=0$ :

R=\frac{2u^2\sin\alpha\cos\alpha}{g}=\frac{u^2\sin2\alpha}{g} \;\Rightarrow\; \frac{g}{2u^2\cos^2\alpha}=\frac{\tan\alpha}{R}.

Step 2 — Range form of the trajectory

Substituting, the path becomes the factored form

y=x\tan\alpha-\frac{\tan\alpha}{R}\,x^2=x\tan\alpha\left(1-\frac{x}{R}\right)=\frac{\tan\alpha}{R}\,x\,(R-x).

This makes the two ground intercepts $x=0$ and $x=R$ explicit.

Step 3 — Impose the wall condition

The particle just clears the wall, so $y=h$ when $x=d$ :

h=\frac{\tan\alpha}{R}\,d\,(R-d)\;\Rightarrow\;\tan\alpha=\frac{hR}{d\,(R-d)}.

Answer

\boxed{\;\tan\alpha=\frac{R\,h}{d\,(R-d)}\quad\Longrightarrow\quad \alpha=\tan^{-1}\!\frac{R\,h}{d\,(R-d)}.\;}