Question

A plane surface is inclined making an angle $\theta$ with the horizontal. From the top of this inclined plane, a bullet is fired with velocity $v$. The maximum possible range of the bullet on the inclined plane is

• A. $\frac{v^2}{g}$
• B. $\frac{v^2}{g(1+\sin \theta )}$
• C. $\frac{v^2}{g(1-\sin \theta )}$
• D. $\frac{v^2}{g(1+\cos \theta )}$

Answer 1

The correct option is C $\frac{v^2}{g(1-\sin \theta )}$

When the bullet is fired from the bottom of the inclined plane.
Resolving the motion parallel and perpendicular to the inclined plane as shown in the diagram,
$v_x=v\cos (\alpha -\theta )v_y=v\sin (\alpha -\theta )a_x=-g\sin \theta a_y=-g\cos \theta$
If $T$ is the time of flight, then
$0=v\sin (\alpha -\theta )T-\frac{1}{2}g\cos \theta T^2$
(Using second equation of motion along Y-axis)
$\begin{array}{}\Rightarrow T=\frac{2v\sin (\alpha -\theta )}{g\cos \theta }& \text{(1)}\end{array}$
If the bullet covers horizontal distance $OB$ in time $T$,
$OB=v\cos \alpha \times T$

Now, $\cos \theta =\frac{OB}{OA}$ (From the figure)
$\Rightarrow OA=\frac{OB}{\cos \theta }=\frac{v\cos \alpha \times T}{\cos \theta }$
Using $\left(1\right)$,
$\Rightarrow OA=\frac{v^2}{g{\cos }^2\theta }\left[2\sin \right(\alpha -\theta \left)\cos \alpha \right]$
Using the identity $2\sin A\cos B=\sin (A+B)+\sin (A-B)$

$OA=\frac{v^2}{g{\cos }^2\theta }\left[\sin \right(2\alpha -\theta )-\sin \theta ]$
Range of a projectile on an inclined plane was found to be
$OA=\frac{v^2}{g{\cos }^2\theta }\left[sin\right(2\alpha -\theta )-\sin \theta ]$

Here $\alpha$ is the angle made by the direction of projection with the horizontal when it is projected from the bottom of the plane.
When it is projected from the top of the plane, the angle made by the projectile with the inclined plane is $\alpha +\theta$.

That is $\theta$ is replaced with $-\theta$ in the above formula. Hence,
$OA=\frac{v^2}{g{\cos }^2\theta }\left[\sin \right(2a-(-\theta ))-\sin (-\theta \left)\right]=\frac{v^2}{g{\cos }^2\theta }\left[\sin \right(2\alpha +\theta )+\sin \theta ]$

$OA$ is maximum when $2\alpha +\theta =\frac{\pi }{2}$ and $\sin (2\alpha +\theta )=1$
$\Rightarrow O{A}_{max}=\frac{v^2}{g{\cos }^2\theta }[1+\sin \theta ]\Rightarrow O{A}_{max}=R_{max}=\frac{v^2}{g(1-{\sin }^2\theta )}[1+\sin \theta ]=\frac{v^2}{g(1-\sin \theta )}$