Question

A plane surface is inclined making an angle $\theta$ with the horizontal. From the bottom 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 B $\frac{v^2}{g(1+\sin \theta )}$

Let the bullet be fired at angle $Ф+\theta$ with the horizontal. The equations of motion of the bullet are:

$x=vcos(Ф+\theta )\times t$

$y=vsin(Ф+\theta )\times t-\frac{1}{2}\times g{t}^2$

or, the trajectory : $y=\left(x\right)tan(Ф+\theta )=\frac{g}{(2v{)}^2}{x}^{2}Se{c}^2(Ф+\theta )$ --- (1)

At the point of the maximum trajectory on the slope (intersection with parabola), we have $y=\left(x\right)tan\theta$ . So (1) becomes

=> $\left(x\right)\left[tan\right(Ф+\theta )-tan\theta ]=\frac{g}{2v^2}\left(x^2\right)se{c}^2(Ф+\theta )$

=> $x=0$ or,

$sin(Ф+\theta )\times cos\theta -cos(Ф+\theta )\times sin\theta =\frac{g}{2v^2}\times se{c}^2(Ф+\theta )cos(Ф+\theta )\times cos\theta$

=> $2sinФ\times cos(Ф+\theta )=\frac{g}{v^2}\left(cos\theta \right)$

=> $x=\left[sin\right(2Ф+\theta )-sin\theta ]\times \left[\frac{v^{2}Sec\theta }{g}\right]$

Range $R$ on the slope = $R=\left(x\right)sec\theta$

=> $R=\left[\frac{v^{2}Se{c}^2\theta }{g}\right)\times \left[Sin\right(2Ф+\theta )-Sin\theta ]$

Maximizing $R$ wrt variable angle $Ф$, means $2Ф+\theta =\frac{\pi }{2}$

=> $Ф=\frac{\pi }{4}-\frac{\theta }{2}$

Then maximum range along the slope

= $\frac{v^{2}Se{c}^2\theta }{g}\times [1-sin\theta ]$

= $\frac{v^2}{g(1+sin\theta )}$