The correct options are
A If the particle strikes the plane at right angles, then
tanα=cotβ+2tanβ B If the particle strikes the plane horizontally, then
tanα=2tanβ
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.
![](https://search-static.byjusweb.com/question-images/byjus/infinitestudent-images/ckeditor_assets/pictures/617533/original_14.png)
The displacement perpendicular to the incline plane is zero, using
2nd equation of motion
y=uyt+12ayt2 0=usin(α−β)t+12(−gcosβ)t2 On rearranging, we get
t=2usin(α−β)gcosβ(1) 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,
vx=ux+axt 0=ucos(α−β)−gsinβ t Substituting value of
t from
(1),
ucos(α−β)=gsinβ2usin(α−β)gcosβ On rearranging,
cot(α−β)=2tanβ(2) Using the identity,
tan(α−β)=tanα−tanβ1+tanα tanβ Equation
(2) becomes,
tanα=cotβ+2tanβ Hence option 1 is correct.
Case 2: When the particle strikes the plane horizontally.
![](https://search-static.byjusweb.com/question-images/byjus/infinitestudent-images/ckeditor_assets/pictures/617520/original_15.png)
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,
vyr=uyr+gt 0=usinα−gt.
Substituting the value of
t from
(1), we get
0=usinα−g2usin(α−β)gcosβ rearranging the above equation we get,
sinα=2sin(α−β)cosβ(3) Using the identity,
sin(α−β)=sinα cosβ−cosα sinβ Equation
(3) becomes,
tanα=2 tanβ Hence option 2 is correct.