Column-I (Angle of projection)Column-IIA. the | vedclass

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

Question

(C) For a projectile,the kinetic energy at the highest point is $K_h = \frac{1}{2}m(u \cos 	heta)^2$ and initial kinetic energy is $K_i = \frac{1}{2}

Vedclass · Accepted Answer

$A-4, B-1, C-3, D-2$

Vedclass · Answer

(C) For a projectile,the kinetic energy at the highest point is $K_h = \frac{1}{2}m(u \cos 	heta)^2$ and initial kinetic energy is $K_i = \frac{1}{2}mu^2$. Thus,$\frac{K_h}{K_i} = \cos^2 	heta$.For $A: 	heta = 45^{\circ}, \frac{K_h}{K_i} = \cos^2(45^{\circ}) = \frac{1}{2}$. (None match directly,let's evaluate $\frac{R}{H} = 4 \cot 	heta$)For $A: 	heta = 45^{\circ}, \frac{R}{H} = 4 \cot(45^{\circ}) = 4$. So $A-4$.For $B: 	heta = 60^{\circ}, \frac{K_h}{K_i} = \cos^2(60^{\circ}) = \frac{1}{4}$. So $B-1$.For $C: 	heta = 30^{\circ}, \frac{R}{H} = 4 \cot(30^{\circ}) = 4\sqrt{3}$. So $C-3$.For $D: 	heta = 	an^{-1}(4), \frac{gT^2}{R} = 2 	an 	heta = 2(4) = 8$. So $D-2$.Matching: $A-4, B-1, C-3, D-2$.

Column-$I$ (Angle of projection)	Column-$II$
$A. \theta = 45^{\circ}$	$1. \frac{K_h}{K_i} = \frac{1}{4}$
$B. \theta = 60^{\circ}$	$2. \frac{gT^2}{R} = 8$
$C. \theta = 30^{\circ}$	$3. \frac{R}{H} = 4\sqrt{3}$
$D. \theta = \tan^{-1}(4)$	$4. \frac{R}{H} = 4$

Similar Questions

At a given instant of time,the position vector of a particle moving in a circle with a velocity $\vec{v} = 3 \hat{i} - 4 \hat{j} + 5 \hat{k}$ is $\vec{r} = \hat{i} + 9 \hat{j} - 8 \hat{k}$. Its angular velocity $\vec{\omega}$ at that time is:

The kinetic energy of a particle moving along a circle of radius $R$ depends on the distance $s$ as $K = as^2$,where $a$ is a constant. Then the force acting on the particle is

$A$ particle has an initial velocity of $(\hat{i} + \hat{j}) \, m/s$ and an acceleration of $(\hat{i} + \hat{j}) \, m/s^2$. Its speed after $10 \, s$ will be:

$A$ particle has an initial velocity $(2\hat i + 3\hat j)$ and an acceleration $(0.3\hat i + 0.2\hat j)$. Its speed after $10\,s$ is:

Two particles $A$ and $B$ start at the origin $O$ and travel in opposite directions along a circular path of radius $0.5\,m$ at constant speeds $0.5\,m/s$ and $1.5\,m/s$,respectively. The time when they collide with each other is ........ $\text{s}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module