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Question

The maximum range of projectile is 500 m. If the particle is thrown up a plane, which is inclined at an angle of 30o with the same speed, the distance covered by it along the inclined plane will be:

A
250 m
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B
500 m
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C
750 m
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D
1000 m
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Solution

The correct option is B 500 m
Range = u2sin2θg ..(1)
and
Rmax=42g ... (ii)
From (1) , max rang at θ=45
500=u2×sin90g
u2g=500
Now,
The distance covered along inclined plane is
v2u2=2as
Here a=gsin30 and v=0
So,
0u2=2(gsin30)s
s=u2g
s=500m


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