Question

A projectile of mass \(m\) is launched from the ground with the initial speed \(v_0\) at an angle \(30^{\circ}\) 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 \(f \times \left(\dfrac{mv_0^{3}}{g}\right)\). The value of \(f\) (rounded off to two decimal places) is .

Hint:

Hint:
Write \(L=T-U\) using the standard projectile \(x(t), y(t)\), and integrate over the whole flight time \(T_f = 2v_0\sin\theta/g\); the closed form comes out to \(S=\dfrac{mv_0^3\sin(3\theta)}{3g}\).

Answer 1

Correct Answer: 0.33

Step 1: Understanding the Concept:
Hamilton's action here is \(S = \int L\, dt\), where the Lagrangian for a projectile with no drag is \(L = T - U = \dfrac{1}{2}m(v_x^2+v_y^2) - mgy\), taking the launch point as the reference for height. We integrate \(L\) over the whole flight, from launch (\(t=0\)) to landing (\(t=T_f\)), for launch angle \(\theta = 30^{\circ}\).

Step 2: Key Formula or Approach:
With \(x,y\) measured from the launch point:
\(v_x = v_0\cos\theta\) (constant), \(\quad v_y = v_0\sin\theta - gt\), \(\quad y(t) = v_0\sin\theta\,t - \dfrac{1}{2}gt^{2}\)
The particle lands when \(y=0\) again, at \(T_f = \dfrac{2v_0\sin\theta}{g}\). Substitute these into \(L\), simplify, then integrate term by term over \(t\) from \(0\) to \(T_f\).

Step 3: Detailed Explanation:
Substituting and expanding:
\[ L = \dfrac{1}{2}m\Big[v_0^2\cos^2\theta + (v_0\sin\theta-gt)^2\Big] - mg\Big(v_0\sin\theta\,t - \dfrac{1}{2}gt^2\Big) \]
\[ L = \dfrac{1}{2}mv_0^{2} - 2mgv_0\sin\theta\,t + mg^{2}t^{2} \]
Integrate each term from \(0\) to \(T_f = \dfrac{2v_0\sin\theta}{g}\):
\[ \int_0^{T_f}\dfrac{1}{2}mv_0^2\,dt = \dfrac{1}{2}mv_0^2T_f = \dfrac{mv_0^3\sin\theta}{g} \]
\[ \int_0^{T_f}\big(-2mgv_0\sin\theta\,t\big)\,dt = -mgv_0\sin\theta\,T_f^2 = -\dfrac{4mv_0^3\sin^3\theta}{g} \]
\[ \int_0^{T_f}mg^2t^2\,dt = \dfrac{1}{3}mg^2T_f^3 = \dfrac{8mv_0^3\sin^3\theta}{g} \]
Adding the three results:
\[ S = \dfrac{mv_0^3}{g}\Big[\sin\theta - 4\sin^3\theta + \dfrac{8}{3}\sin^3\theta\Big] = \dfrac{mv_0^3}{3g}\big(3\sin\theta - 4\sin^3\theta\big) \]
Since \(3\sin\theta - 4\sin^3\theta = \sin(3\theta)\):
\[ S = \dfrac{mv_0^3\sin(3\theta)}{3g} \]
At \(\theta = 30^{\circ}\), \(3\theta = 90^{\circ}\) and \(\sin(90^{\circ}) = 1\), so \(S = \dfrac{mv_0^3}{3g}\).

Step 4: Final Answer:
Comparing with \(S = f\times\dfrac{mv_0^3}{g}\) gives \(f = \dfrac{1}{3} = 0.3333\ldots\), which rounds to \(0.33\). \[ \boxed{f = 0.33} \]