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Question

A particle is projected up an inclined plane of inclination β at an elevation α to the horizon. Which of the following are true ?

A
If the particle strikes the plane at right angles, then tanα=cotβ+2tanβ
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B
If the particle strikes the plane horizontally, then tanα=2tanβ
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C
If the particle strikes the plane at right angles, then tanβ=cotβ+2tanα
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D
If the particle strikes the plane horizontally, then tanβ=2tanα
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Solution

The correct option is 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.

The displacement perpendicular to the incline plane is zero, using 2nd equation of motion
y=uyt+12ayt2
0=usin(αβ)t+12(gcosβ)t2
On rearranging, we get
t=2usin(αβ)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(αβ)gsinβ t
Substituting value of t from (1),
ucos(αβ)=gsinβ2usin(αβ)gcosβ
On rearranging,
cot(αβ)=2tanβ(2)
Using the identity, tan(αβ)=tanαtanβ1+tanα tanβ
Equation (2) becomes,
tanα=cotβ+2tanβ
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,
vyr=uyr+gt
0=usinαgt.
Substituting the value of t from (1), we get
0=usinαg2usin(αβ)gcosβ
rearranging the above equation we get,
sinα=2sin(αβ)cosβ(3)
Using the identity,
sin(αβ)=sinα cosβcosα sinβ
Equation (3) becomes,
tanα=2 tanβ
Hence option 2 is correct.

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