GATE 2026 PH – Question 28
A projectile of mass $m$ is launched from the ground with the initial speed $v_0$ at an angle 30° from the horizontal. Take the ground to be horizontal. Ignoring the drag, the magnitude of Hamilton’s action $\int L\,dt$ for the particle from the beginning till it hits the ground is $f\times(mv_0^3/g)$. The value of $f$ (rounded off to two decimal places) is _____
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: 0.32 to 0.34
Explanation
**Action** for the projectile: $S=\int_0^{t_f}L\,dt$ with $L=T-V=\tfrac12m\dot x^2+\tfrac12m\dot y^2-mgy$.
**Motion** (no drag): $\dot x=v_0\cos\theta$, $\dot y=v_0\sin\theta-gt$, $y=v_0\sin\theta\,t-\tfrac12gt^2$. The flight time is $t_f=\dfrac{2v_0\sin\theta}{g}$.
**Lagrangian along the path:**
$$L=\tfrac12m\left[v_0^2\cos^2\theta+(v_0\sin\theta-gt)^2\right]-mg\left(v_0\sin\theta\,t-\tfrac12gt^2\right)$$
$$=\tfrac12mv_0^2-2mgv_0\sin\theta\,t+mg^2t^2 .$$
**Integrate from 0 to $t_f$:**
$$S=\tfrac12mv_0^2t_f-mgv_0\sin\theta\,t_f^2+\tfrac13mg^2t_f^3 .$$
With $t_f=\dfrac{2v_0\sin\theta}{g}$:
$$S=\frac{mv_0^3}{g}\left[\sin\theta-4\sin^3\theta+\tfrac83\sin^3\theta\right]=\frac{mv_0^3}{g}\left(\sin\theta-\tfrac43\sin^3\theta\right).$$
**At $\theta=30^\circ$:** $\sin\theta=\tfrac12$, so $S=\dfrac{mv_0^3}{g}\left(\tfrac12-\tfrac43\cdot\tfrac18\right)=\dfrac{mv_0^3}{g}\left(\tfrac12-\tfrac16\right)=\dfrac13\dfrac{mv_0^3}{g}$.
$$f=\mathbf{0.33}.$$