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

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

Column $I$	Column $II$
$(A)$ Relative acceleration between two	$(p)$ $0$ $SI$ unit
$(B)$ Relative velocity between two	$(q)$ $5$ $SI$ unit
$(C)$ Horizontal distance between two	$(r)$ $10$ $SI$ unit
$(D)$ Vertical distance between two	$(s)$ $20$ $SI$ unit

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