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UPSC 2021 Maths Optional Paper 1 Q8b — Step-by-Step Solution

15 marks · Section B

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

Question

Describe motion and path of projectile from $(0,0)$ with velocity $u$ at angle $\theta$ to horizontal. If particles are projected from same point in same vertical plane with velocity $4\sqrt g$, determine locus of vertices of their paths.

Technique

Standard projectile parametric form; vertex height/range as function of $\theta$; eliminate $\theta$ to get locus.

Solution

Part 1 — Projectile motion description

Standard parametric form with $x$ horizontal, $y$ vertical (up positive):

$x(t)=ut\cos\theta$, $y(t)=ut\sin\theta-\tfrac{1}{2}gt^2$.

Eliminate $t=x/(u\cos\theta)$: $y=x\tan\theta-\dfrac{gx^2}{2u^2\cos^2\theta}=x\tan\theta-\dfrac{gx^2\sec^2\theta}{2u^2}$.

This is a parabola opening downward, with vertex at the highest point.

Height of vertex: $H=\dfrac{u^2\sin^2\theta}{2g}$. $x$-coordinate of vertex: $X=\dfrac{u^2\sin 2\theta}{2g}=\dfrac{u^2\sin\theta\cos\theta}{g}$.

Part 2 — Locus of vertices (vary $\theta$, fix $u$)

With $u=4\sqrt g$, $u^2=16g$.

$X=\dfrac{16g\sin\theta\cos\theta}{g}=16\sin\theta\cos\theta=8\sin 2\theta$.

$Y=H=\dfrac{16g\sin^2\theta}{2g}=8\sin^2\theta=4(1-\cos 2\theta)$.

Step 3 — Eliminate $\theta$

Let $\phi=2\theta$. Then $X=8\sin\phi$, $Y=4(1-\cos\phi)$, i.e., $\cos\phi=1-Y/4$, $\sin\phi=X/8$.

$\sin^2\phi+\cos^2\phi=1$: $(X/8)^2+(1-Y/4)^2=1$, $\dfrac{X^2}{64}+\!\left(1-\dfrac{Y}{4}\right)^2=1$.

Expand: $\dfrac{X^2}{64}+1-\dfrac{Y}{2}+\dfrac{Y^2}{16}=1$, $\dfrac{X^2}{64}+\dfrac{Y^2}{16}=\dfrac{Y}{2}$.

Multiply by 64: $X^2+4Y^2=32Y$.

Rearrange: $X^2+4Y^2-32Y=0$, $X^2+4(Y^2-8Y)=0$, $X^2+4(Y-4)^2-64=0$, $X^2+4(Y-4)^2=64$, $\dfrac{X^2}{64}+\dfrac{(Y-4)^2}{16}=1$.

This is an ellipse centred at $(0,4)$ with semi-axes $8$ (horizontal) and $4$ (vertical).

Answer