A projectile is fired from the surface of the | vedclass

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(A) The equation of the trajectory of a projectile is given by $y = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	heta}$.For two trajectories to be iden

Vedclass · Accepted Answer

$3.5$

Vedclass · Answer

(A) The equation of the trajectory of a projectile is given by $y = x 	an 	heta - \frac{g x^2}{2 u^2 \cos^2 	heta}$.For two trajectories to be identical,the coefficients of $x$ and $x^2$ must be the same.Since the angle $	heta$ is the same,$	an 	heta$ is constant.For the $x^2$ term to be identical,$\frac{g_e}{u_e^2} = \frac{g_p}{u_p^2}$,where $g_e$ and $u_e$ are gravity and velocity on Earth,and $g_p$ and $u_p$ are gravity and velocity on the planet.Given $u_e = 5\, ms^{-1}$,$u_p = 3\, ms^{-1}$,and $g_e = 9.8\, ms^{-2}$.Substituting the values: $\frac{9.8}{5^2} = \frac{g_p}{3^2}$.$g_p = 9.8 	imes \frac{9}{25} = 9.8 	imes 0.36 = 3.528\, ms^{-2}$.Rounding to the nearest provided option,we get $3.5\, ms^{-2}$.

List-$I$	List-$II$
$(i) P$	$(A) (0, 0)$
$(ii) Q$	$(B) (6, 0)$
$(iii) R$	$(C) (-2, 1)$
$(iv) S$	$(D) (-6, 0)$
	$(E) (-6, -3)$
	$(F) (-6, 3)$

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