The trajectory of a particle in projectile mo | vedclass

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Question

(A) The standard equation of a projectile trajectory is $y = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	heta}$.Comparing $y = x - \frac{x^2}{80}$ wit

Vedclass · Accepted Answer

$(A \rightarrow r, B \rightarrow s, C \rightarrow p, D \rightarrow q)$

Vedclass · Answer

(A) The standard equation of a projectile trajectory is $y = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	heta}$.Comparing $y = x - \frac{x^2}{80}$ with the standard equation:$(A)$ $	an 	heta = 1 \Rightarrow 	heta = 45^{\circ}$. Thus, $(A ightarrow r)$.$(B)$ The horizontal range $R = 80 \, m$. Since $y = x(1 - x/80)$, $y=0$ at $x=0$ and $x=80 \, m$. The time of flight $T = \frac{2 u \sin 	heta}{g}$. From $\frac{g}{2 u^2 \cos^2 	heta} = \frac{1}{80}$, we get $u^2 = \frac{80 	imes 10}{2 \cos^2 45^{\circ}} = 800 \Rightarrow u = 20\sqrt{2} \, m/s$. $T = \frac{2 	imes 20\sqrt{2} 	imes (1/\sqrt{2})}{10} = 4 \, s$. At $t=4 \, s$, the particle is at the ground, and the angle of velocity with the horizontal is $-45^{\circ}$ (or $45^{\circ}$ below horizontal). Given the options, $(B ightarrow r)$ is the intended match.$(C)$ Maximum height $H = \frac{u^2 \sin^2 	heta}{2g} = \frac{800 	imes (1/2)}{20} = 20 \, m$. Thus, $(C ightarrow p)$.$(D)$ Horizontal range $R = 80 \, m$. Thus, $(D ightarrow q)$.Therefore, the correct match is $(A ightarrow r, B ightarrow r, C ightarrow p, D ightarrow q)$.

$Column-I$	$Column-II$
$(A)$ Angle of projection	$(p)$ $20 \, m$
$(B)$ Angle of velocity with horizontal after $4 \, s$	$(q)$ $80 \, m$
$(C)$ Maximum height	$(r)$ $45^{\circ}$
$(D)$ Horizontal range	$(s)$ $\tan^{-1}(1/2)$

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