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

A
ut3
B
3ut2
C
3ut
D
2ut

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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