Question

A stone is projected from point A with speed u making an angle ${60}^0$ with horizontal as shown. The fixed inclined surface makes an angle ${30}^0$ with horizontal. The stone lands at B after time 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

The correct option is A $\frac{ut}{\sqrt{3}}$

since the velocity of projected is u and inclination is ${60}^0$

then,

Its, velocity along $A{B}^{\prime }=u\cos {60}^0$

$=u\times \frac{1}{2}$

$=\frac{u}{2}$

Time taken to reach at $B=t$ (given)

It same time ${}^{\prime }{t}^{\prime }$ it travels horizontal distance $A{B}^{\prime }$

$\Rightarrow A{B}^{\prime }=\frac{u}{2}\times t$

From $△AB{B}^{\prime }$

$AB\cos {30}^0=A{B}^{\prime }$

$AB\times \frac{\sqrt{3}}{2}=\frac{ut}{2}$

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