Intra-Ring Bombardment

Classical Mechanics Level pending

Define two points as follows:

$\large{\vec{P}_1 = (P_{1x},P_{1y},P_{1z}) = (1,0,0) \\ \vec{P}_2 = (P_{2x},P_{2y},P_{2z}) = \Big(\cos (\theta),\sin (\theta),0 \Big) }$

Gravity is $10 \, \text{m/s}^2$ in the negative $z$ direction. A projectile is launched from point $\vec{P}_1$ with speed $v$ at an angle of $\pi/4$ with respect to the $xy$ plane, such that it lands again in the $xy$ plane at point $\vec{P}_2$ . The value of $v$ therefore varies with $\theta$ .

If $\theta$ is chosen randomly and uniformly within the range $0 \leq \theta \leq 2 \pi$ , what is the expected value of the launch speed $v$ ?

The answer is 3.411.

1 solution

Karan Chatrath
Jun 8, 2019

The points of launch and landing are given. This can be used to compute the range of the projectile. $R = \sqrt{(P_{1x} - P_{2x})^2 + (P_{1y} - P_{2y})^2} = 2\sin\bigg(\frac{\theta}{2}\bigg)$

The range of the projectile for a launch angle of 45 degrees is also given by: $R = \frac{v^2}{g}$

Equating these two expression gives launch speed as a function of $\theta$ .

$v = \sqrt{2g\sin\bigg(\frac{\theta}{2}\bigg)}$

The expected value of launch speed is then computed:

$v_{exp} = \frac{1}{2\pi}\int_{0}^{2\pi}vd\theta$

Which is (after trivial simplification):

$v_{exp} = \frac{2\sqrt{2g}}{\pi}\int_{0}^{\pi/2}\sqrt{\sin(x)}dx$

This integral can be solved using the concept of Beta or Gamma functions. The answer is: 3.411 .

Intra-Ring Bombardment

The answer is 3.411.

1 solution

0 pending reports