When an object is shot from the bottom of a l | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Using the third equation of motion,$v^{2} = u^{2} - 2as$. At the highest point,the final velocity $v = 0$.Therefore,$0 = u^{2} - 2as$,which gives

Vedclass · Accepted Answer

$1: \sqrt{3}$

Vedclass · Answer

(C) Using the third equation of motion,$v^{2} = u^{2} - 2as$. At the highest point,the final velocity $v = 0$.Therefore,$0 = u^{2} - 2as$,which gives $s = \frac{u^{2}}{2a}$.For an object on a smooth inclined plane,the acceleration $a = g \sin 	heta$.Thus,the distance traveled is $s = \frac{u^{2}}{2g \sin 	heta}$.Since $u$ and $g$ are constant,$s \propto \frac{1}{\sin 	heta}$.Therefore,$\frac{x_{1}}{x_{2}} = \frac{\sin 	heta_{2}}{\sin 	heta_{1}} = \frac{\sin 30^{\circ}}{\sin 60^{\circ}}$.Substituting the values,$\frac{x_{1}}{x_{2}} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}$.Hence,the ratio $x_{1}: x_{2}$ is $1: \sqrt{3}$.

Similar Questions

$A$ cube of mass $m$ slides down an inclined right-angle trough. If the coefficient of kinetic friction between the cube and the trough is $\mu_k$,then the acceleration of the block is:

$A$ block of mass $m$ is placed on a surface having a vertical cross-section given by $y = x^2 / 4$. If the coefficient of friction is $0.5$,the maximum height above the ground at which the block can be placed without slipping is:

The tension $T$ in the string shown in the figure is:

An ice cube is kept on an inclined plane of angle $30^{\circ}$. The coefficient of kinetic friction between the block and the inclined plane is $(1 / \sqrt{3})$. What is the acceleration of the block? $...... \ m / s ^2$

$A$ cubic block of mass $m$ is sliding down on an inclined plane at $60^{\circ}$ with an acceleration of $\frac{g}{2}$. The value of the coefficient of kinetic friction is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module