Question

A particle is projected up an inclined plane of inclination $\beta$ at an elevation $\alpha$ to the horizon. Which of the following are true ?

• A. If the particle strikes the plane at right angles, then $\tan \alpha =\cot \beta +2\tan \beta$
• B. If the particle strikes the plane horizontally, then $\tan \alpha =2\tan \beta$
• C. If the particle strikes the plane at right angles, then $\tan \beta =\cot \beta +2\tan \alpha$
• D. If the particle strikes the plane horizontally, then $\tan \beta =2\tan \alpha$

Answer 1

Answer: (A), (B)

If the particle strikes the plane horizontally, then $\tan \alpha =2\tan \beta$

Case 1: When the particle strikes the plane at right angle.
Let the x-axis along the incline plane and y-axis normal to the incline plane.

The displacement perpendicular to the incline plane is zero, using $2^{nd}$ equation of motion
$y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$0=usin(\alpha -\beta )t+\frac{1}{2}(-gcos\beta )t^2$
On rearranging, we get
$\begin{array}{}t=\frac{2usin(\alpha -\beta )}{gcos\beta }& \text{(1)}\end{array}$
When the ball strikes the pane, x-component of velocity or the component of velocity along the inclined plane will be zero. Thus using 1st equation of motion,
$v_x=u_x+a_{x}t$
$0=ucos(\alpha -\beta )-gsin\beta t$
Substituting value of $t$ from $\left(1\right)$,
$ucos(\alpha -\beta )=gsin\beta \frac{2usin(\alpha -\beta )}{gcos\beta }$
On rearranging,
$\begin{array}{}cot(\alpha -\beta )=2tan\beta & \text{(2)}\end{array}$
Using the identity, $tan(\alpha -\beta )=\frac{tan\alpha -tan\beta }{1+tan\alpha tan\beta }$
Equation $\left(2\right)$ becomes,
$\tan \alpha =\cot \beta +2\tan \beta$
Hence option 1 is correct.
Case 2: When the particle strikes the plane horizontally.
In terms of regular coordinate system i.e when x-axis is horizontal and y-axis is vertical, vertcal component of velocity will be zero.
Hence applying the 1st equation of motion in normal coordinate axis,
$v_{yr}=u_{yr}+gt$
$0=usin\alpha -gt$.
Substituting the value of $t$ from $\left(1\right)$, we get
$0=usin\alpha -g\frac{2usin(\alpha -\beta )}{gcos\beta }$
rearranging the above equation we get,
$\begin{array}{}sin\alpha =\frac{2sin(\alpha -\beta )}{cos\beta }& \text{(3)}\end{array}$
Using the identity,
$sin(\alpha -\beta )=sin\alpha cos\beta -cos\alpha sin\beta$
Equation $\left(3\right)$ becomes,
$tan\alpha =2tan\beta$
Hence option 2 is correct.