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
B
500m
C
750m
D
1000m

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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