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UPSC 2015 Maths Optional Paper 1 Q7b — Step-by-Step Solution

13 marks · Section B

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

Question

A particle is projected from the base of a hill whose slope is that of a right circular cone, whose axis is vertical. The projectile grazes the vertex and strikes the hill again at a point on the base. If the semivertical angle of the cone is $30^\circ$ and $h$ is the height, determine the initial velocity $u$ of the projection and its angle of projection.

Technique

Standard projectile motion with two conditions: max-height $=h$ (apex of cone) and range $=2R=2h/\sqrt 3$. Two equations, two unknowns $u,\alpha$.

Solution

Setup. Right-circular cone with vertical axis, apex at height $h$, semivertical angle $30^\circ$. Base radius $R=h\tan 30^\circ=h/\sqrt 3$.

The trajectory of the projectile is a parabola in some vertical plane. By the symmetry (cone is rotationally symmetric, the trajectory crosses the axis at the apex), the parabola must lie in a vertical plane containing the cone’s axis.

Set 2D coordinates in this plane: $x$ horizontal (along the base diameter), $z$ vertical. The cone’s profile in this plane is two slanted lines from the apex $(0,h)$ to the base points $(\pm R,0)=(\pm h/\sqrt 3,0)$.

The projectile is launched from some point on the base, grazes the apex $(0,h)$, and lands at another point on the base.

Question: Where exactly does it launch from? The base of the cone is a circle of radius $R$. Symmetry: if it lands on the base and grazes the apex, and the apex is on the axis, then the trajectory passes through the apex. The two impact points (launch + landing) must be symmetric about the apex’s vertical line — i.e. at $(-R,0)$ and $(+R,0)$.

So launch at $(-R,0)$, apex of trajectory at $(0,h)$, landing at $(R,0)$.

The vertex (apex of cone) is the highest point of the trajectory. “Grazes” means the trajectory just touches the apex — i.e. the apex is the maximum of the parabolic trajectory. (Mathematically, “graze” means the trajectory passes through and has the cone’s surface as tangent at that point.)

Step 1 — Trajectory passes through $(-R,0)$, $(0,h)$, $(R,0)$

A parabolic trajectory with horizontal axis… no, the trajectory $z(x)$ has the form $z=$ linear$\cdot x$ + $x^2$-term. Launch at $(-R,0)$ with speed $u$ at angle $\alpha$ above horizontal:

$x(t)=-R+(u\cos\alpha)t,\;z(t)=(u\sin\alpha)t-\tfrac{1}{2}gt^2.$

Trajectory $z$ as function of $x$: let $\xi=x-(-R)=x+R$, so $\xi=(u\cos\alpha)t\Rightarrow t=\xi/(u\cos\alpha)$.

At apex $\xi=R$ (i.e. $x=0$): $z=h$. At landing $\xi=2R$ (i.e. $x=R$): $z=0$.

Step 2 — Apex of trajectory is at $(0,h)$

Symmetric parabola: launch at $\xi=0$, landing at $\xi=2R$, apex (highest point) at $\xi=R$.

Maximum height of a projectile launched at angle $\alpha$ with speed $u$:

Set $H_{\max}=h$:

Step 3 — Range of projectile = $2R$

Range on level ground:

Set Range $=2R=2h/\sqrt 3$:

Step 4 — Divide (1) by (2)

Step 5 — Find $u$

From $\tan\alpha=2\sqrt 3$: $\sin\alpha=\dfrac{2\sqrt 3}{\sqrt{1+12}}=\dfrac{2\sqrt 3}{\sqrt{13}}$, $\cos\alpha=\dfrac{1}{\sqrt{13}}$.

Substitute in (1): $u^2\cdot\dfrac{12}{13}=2gh\Rightarrow u^2=\dfrac{26gh}{12}=\dfrac{13gh}{6}$.

Step 6 — Verify the “grazes” condition

“Grazes the vertex” — for an ordinary parabolic trajectory, the apex is just the highest point, and the trajectory has a horizontal tangent there. For it to graze the cone’s apex, we need the parabola to touch the cone’s surface there.

The cone’s vertical-plane profile near $(0,h)$ consists of two slanted lines meeting at the apex. The trajectory has horizontal tangent at $(0,h)$. So the trajectory’s tangent is horizontal, and the cone’s surfaces have slopes $\pm\cot 30^\circ=\pm\sqrt 3$ at the apex (slopes of the two slanted sides).

“Graze” = touches the apex without crossing the cone’s interior. Since the trajectory is horizontal at the apex while the cone slopes downward on both sides at $\pm\sqrt 3$, the trajectory will immediately exit the cone (going outside). So actually the trajectory passes through the apex but does not stay in/on the cone — this is consistent with the projectile flying outside the cone for most of the trajectory.

Hmm — but the problem says it lands on the base. Let me reconsider.

Actually, the projectile is outside the cone — it’s flying in open air, and the cone is a hill it must clear. It starts on the base (at the edge of the cone, at $(-R,0)$), rises up (outside the cone), grazes the apex $(0,h)$, then descends on the other side back to the base.

For it to “graze” — the trajectory must be tangent to the cone’s surface at the apex from outside. The cone’s surface at the apex has two slopes (a ridge). For the trajectory (horizontal at apex) to be tangent, we need… hmm, actually the cone is a single point at the apex, so “graze” simply means touch.

But wait — for the trajectory not to cross the cone’s side surfaces on the way up, it must be entirely above (outside) the cone’s slant surface, except at the apex.

So the trajectory $z(\xi)=\tan\alpha\cdot\xi-\dfrac{g\xi^2}{2u^2\cos^2\alpha}$ must satisfy $z(\xi)\ge$ (cone surface height at horizontal position $x=-R+\xi$) for all $\xi\in[0,2R]$.

Cone surface in this plane: for $x\in[-R,0]$, the slant from $(-R,0)$ to $(0,h)$ has equation $z=h(x+R)/R=h(1+x/R)$ — wait we’re inside the cone… Let me redo.

For $x\in[-R,R]$, the cone’s surface in the vertical plane is a line from $(-R,0)$ up to $(0,h)$ (left slant) then back down to $(R,0)$ (right slant):

Left slant: $z=h\cdot(x+R)/R$ for $x\in[-R,0]$. Right slant: $z=h\cdot(R-x)/R$ for $x\in[0,R]$.

The trajectory must be above these for all $x$. At launch ($x=-R$, $z=0$) — equal. At apex ($x=0$, $z=h$) — equal. At landing ($x=R$, $z=0$) — equal. In between, trajectory must be above the slant.

Tangency at the launch point — the trajectory tangent at $(-R,0)$ should equal or exceed the slant’s slope ($h/R$) for the trajectory not to dip below the slant immediately.

Launch tangent slope $=\tan\alpha$. Slant slope (left) $=h/R=h/(h/\sqrt 3)=\sqrt 3$.

We found $\tan\alpha=2\sqrt 3$, which is steeper than $\sqrt 3$ ✓. So the trajectory rises above the left slant immediately — good.

Similarly the tangent at landing has slope $-\tan\alpha=-2\sqrt 3$, steeper (in magnitude) than the right slant $-\sqrt 3$ — also above ✓.

The condition “grazes” simply requires the trajectory to touch the cone at the apex but not anywhere else along the slant — which is automatic here since the parabola’s max is exactly at the apex.

Summary

Answer