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The maximum range of a projectile is 500m. If the particle is thrown up a plane is inclined at an angle of `30^(@)` with the same speed, the distance covered by it along the inclined plane will be:
A. 250 m
B. 500 m
C. 750m
D. 1000 m

1 Answer

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Best answer
Correct Answer - B
b. For the maximum range, `theta= 45^@`.
`R = (u^2sin2theta)/g = u^2/g sin 90^@ = u^2/g or 500 = u^2/g`
The distance covered along the inclined plane can be obtained
using the equation,
`v^2-u^2 = 2as`
or `0-u^2 = 2(-g sin 30^@)s or s = u^2/g = 500m` .

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