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Question

A projectile has a maximum range of $$500  m$$. If the projectile is now thrown up an inclined plane of $${ 30 }^{ \circ  }$$ with the same velocity, the distance covered by it along the inclined plane will be about


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

The correct option is B $$500  m$$
we know that maximum range of any object projected from an inclined plain is given as,
$${R}_{max}=\dfrac{{V}^{2}}{g(1+sin{\theta})}$$

so from here we will find the value of $${V}$$ in meter per second,now the second projectile is projected with same velocity therefore we can again use this velocity for the second case to find the value of range.
so, $${R}_{max}=\dfrac{{(86.60)}^{2}}{10(1+sin{30})}={500}\ meter$$

Physics

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