A gun is fired on a horizontal plane whose bullet passes over a wall of height h, where the plane of trajectory is perpendicular to the wall. Given that the firing velocity is $\sqrt{2ga}$ and the angle of projection cannot exceed $\theta$, where $\theta < 45$, prove that the gun can be fired from any point in a strip of the plane of width $4\cos(\theta)\sqrt{(a^2\sin^2\theta-ah)}$
Projectile Motion - Overcoming a wall
-1
$\begingroup$
classical-mechanics
1 Answers
1
Hint...you can obtain the equation of the trajectory which in this case is of the form $$y=x\tan\theta-\frac{x^2}{4a}\sec^2\theta$$
Put $y=h$ and solve this quadratic for $x$. The difference in the two solutions gives you the expression you are looking for.