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Question

A particle is projected with certain velocity at an angle α above the horizontal from the foot of an inclined plane having inclination of 30. If the particle strikes the plane normally then α is:

A
α=30+tan123
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B
45
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C
60
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D
α=30+tan132
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Solution

The correct option is D α=30+tan132
If the projectile hits the inclined plane normally, it means that component of final velocity (v) along the incline is zero i.e vx=0
Also,
ux=ucos(α30)
uy=usin(α30)
ax=gsinθ=gsin30, ay=gcosθ=gcos30

Resolving the whole motion as 2 separate independent 1-D motion along x and y direction respectively, and applying kinematic equation:

vx=ux+axt

0=ucos(α30)gsin30t
t=ucos(α30)gsin30......i

Now, time of flight for a projectile on inclined plane is given by the formula:

T=2uy|ay|=2usin(α30)gcos30..........ii

Equating two expressions (i) and (ii) as both represent the time of flight for particle projected on the inclined plane, we get:

T=t

2usin(α30)gcos30=ucos(α30)gsin30
tan(α30)=12tan30=32

α30=tan132
Hence, α=30+tan132

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