Question

A plane surface is inclined making an angle $\beta$ above the horizon. A bullet is fired with the point of projection at the bottom of the inclined plane with a velocity $u$ ; then the maximum range is given by:

• A. $\frac{u^2}{g}$
• B. $\frac{u^2}{g\left(1+sin\beta \right)}$
• C. $\frac{u^2}{g\left(1-sin\beta \right)}$
• D. $\frac{u^2}{g\left(1+cos\beta \right)}$

Answer 1

Answer: (B)

Suppose the particle be projected with a velocity $u$ making an angle $\theta$ with the horizontal. Suppose the particle strikes the inclined plane at A after time $T$. If $x$-axis is taken along the inclined plane and y-axis normal to the inclined plane. Then



$u_x=u$ cos $(\theta -\beta )$,

$u_y=u$ sin$(\theta -\beta )$

During the time of flight $T$, the displacement along y-axis is zero.

Using equation

$s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$0=u$ sin $(\theta -\beta )T+\frac{1}{2}(-gcos\beta )T^2$

$T=\frac{2usin(\theta -\beta )}{gcos\beta }$

Component of velocity along horizontal $=u$ cos $\theta$.

Distance covered along horizontal

$OB=(ucos\beta )T$

$=ucos\theta [\frac{2usin(\theta -\beta )}{gcos\beta }]$

$=\frac{2u^{2}cos\theta sin(\theta -\beta )}{gcos\beta }$

Range $R=OA$

$OA=\frac{OB}{cos\beta }$

$=\frac{2u^{2}cos\theta \sin (\theta -\beta )}{gco{s}^2\theta }$

$R=\frac{2u^{2}cos\theta sin(\theta -\beta )}{gco{s}^2\beta }$

$=[\frac{u^2}{gco{s}^2\beta }][2sin(\theta -\beta )cos\theta ]$

$=\frac{u^2}{gco{s}^2\beta }[\sin (2\theta -\beta )-\sin \beta ]$
For a given u and $\beta ,R$ is maximum, when sin $(2\theta -\beta )$ is maximum, i.e., when

sin $(2\theta -\beta )=1or2\theta -\beta =\frac{\pi }{2}$

or $2\theta =\beta +\frac{\pi }{2}$ or $\theta =\frac{\beta }{2}+\frac{\pi }{2}$

Hence, the angle of projection $\theta$, for maximum range up the inclined plane is given by

$\theta =\frac{\beta }{2}+\frac{\pi }{4}$

$R_{max}=\frac{u^2}{gco{s}^2\beta }[1-sin\beta ]$

$=\frac{u^2(1-sin\beta )}{g(1-si{n}^2\beta )}$

$=\frac{u^2}{g(1+sin\beta )}$