Question

If $R_1$ and $R_2$ are the maximum ranges of a projectile when it is projected from the bottom and top respectively of an inclined plane respectively, then find the maximum range of the projectile when it is projected from the ground with same velocity.

• A. $\frac{R_1+R_2}{2}$
• B. $\frac{R_1{R}_2}{R_1+R_2}$
• C. $\frac{2R_1{R}_2}{R_1+R_2}$
• D. $\frac{R_1+R_2}{R_1{R}_2}$

Answer 1

The correct option is C $\frac{2R_1{R}_2}{R_1+R_2}$

When the projectile is from top to 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 projected object 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 $2\sin A\cos B=\sin (A+B)+\sin (A-B)$

$OA=R_1=\frac{v^2}{g{\cos }^2\theta }\left[\sin \right(2\alpha -\theta )-\sin \theta ]$
Clearly, the range $R_1(=OA)$ will be maximum when $\sin (2\alpha -\theta )$ is maximum, i.e., 1.
This would mean $2\alpha -\theta =\frac{\pi }{2}\text{or}\alpha =\frac{\theta }{2}+\frac{\pi }{4}$

Maximum range on the inclined plane from bottom to top can be denoted by $R_1$.
$\therefore R_1=\frac{v^2}{g{\cos }^2\theta }(1-\sin \theta )=\frac{v^2(1-\sin \theta )}{g(1-{\sin }^2\theta )}=\frac{v^2(1-\sin \theta )}{g(1+\sin \theta )(1-\sin \theta )}=\frac{v^2}{g(1+\sin \theta )}$

When the projectile is from top to bottom of the inclined plane.
Range of a projectile when it is from bottom of 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 $\sin (2\alpha +\theta )=1$
$\Rightarrow O{A}_{max}=\frac{v^2}{g{\cos }^2\theta }[1+\sin \theta ]\Rightarrow O{A}_{max}=R_2=\frac{v^2}{g(1-{\sin }^2\theta )}[1+\sin \theta ]R_2=\frac{v^2}{g(1-\sin \theta )}$
We know that the maximum range of a projectile from the bottom and top of an inclined plane are, respectively
$R_1=\frac{v^2}{g(1+\sin \theta )}$ and
$R_2=\frac{v^2}{g(1-\sin \theta )}$

Then, $\frac{1}{R_1}+\frac{1}{R_2}=\frac{g(1+\sin \theta )}{v^2}+\frac{g(1-\sin \theta )}{v^2}$
$\Rightarrow \frac{R_1+R_2}{R_1{R}_2}=\frac{2g}{v^2}\Rightarrow \frac{2R_1{R}_2}{R_1+R_2}=\frac{v^2}{g}$

Maximum range of a projectile on a horizontal surface $R_{max}=\frac{v^2}{g}$
Thus, $R_{max}=\frac{2R_1{R}_2}{R_1+R_2}$