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. $\frac{ut}{\sqrt{3}}$
• B. $\frac{\sqrt{3}ut}{2}$
• C. $\sqrt{3}ut$
• D. $2ut$

Answer 1

Answer: (A)

$\frac{ut}{\sqrt{3}}$

we have,

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

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

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