Question

Two balls are thrown from an inclined plane at angle of projection $\alpha$ with the plane one up the incline plane and other down the incline as shown in the figure. If $R_1$ & $R_2$ be their respective ranges, then
[here $T_1$ & $T_2$ are times of flight in the two cases respectively]

• A. $h_1=h_2$
• B. $R_2-R_1=T_1^2$
• C. $R_2-R_1=gsin\theta T_2^2$
• D. $R_2-R_1=gsin\theta T_1^2$

Answer 1

Answer: (A), (C), (D)

$h_1=h_2$
$R_2-R_1=gsin\theta T_2^2$
$R_2-R_1=gsin\theta T_1^2$

If the particle is projected with a velocity v at an angle $\alpha$ with the plane, then the maximum displacement of the particle normal to the plane is given as $h=\frac{(vsin\alpha {)}^2}{2gcos\theta }$
Here $gcos\theta$ is the component of gravitational acceleration normal to the plane.
Since in the both the given cases the particle is projected at an angle $\alpha$ with the plane and the plane has an inclination $\theta$ with horizontal, then both $h_1$ and $h_2$ will be equal to each other and their magnitude will be equal to $\frac{v^{2}si{n}^2\alpha }{2gcos\theta }$.
The time of flight in both the cases will be equal to $\frac{2vsin\alpha }{gcos\theta }$.
Here $vsin\alpha$ is the normal component of projection.
If we consider the motion of particles along the plane then the range of the first particle will be
$R_1=vcos\alpha T-\frac{1}{2}gsin\theta T^2$
and the range of the other particle will be
$R_2=vcos\alpha T+\frac{1}{2}gsin\theta T^2$
Thus we get the value of $R_2-R_1=gsin\theta T^2$
As the value for time of periods for both the particles are same we get both C and D as correct.