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Question

In figure shown below, the time taken by the projectile to reach from $$A$$ to $$B$$ is $$t$$ then, the distance $$AB$$ is equal to 
1229834_c3eb1c9f285c49cbbde97240f661064e.png


A
ut3
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B
3ut2
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C
3ut
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D
2ut
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Solution

The correct option is B $$\dfrac {ut}{\sqrt {3}}$$
we have,

$$AB=AC \sec 30^{\circ}$$

$$=\dfrac{u}{2}t\times \dfrac{2}{\sqrt 3}$$

$$=\dfrac{ut}{\sqrt 3}$$

Physics

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