Time taken by the projectile to reach from A to B is t. Then... | PHYSICS - PROJECTILE MOTION | SolveeIt | Solveeit

Today, August 19

Question

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

2 ut

Answer

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

Explanation

Solution

Horizontal component of velocity

u_H = u cos 60⁰ = $\frac{u}{2}$

\ AC = (u_H)t = $\frac{ut}{2}$

and AB = AC sec 30⁰ = $\left( \frac{ut}{2} \right)$ = $\left( \frac{2}{\sqrt{3}} \right)$ = $\frac{ut}{\sqrt{3}}$