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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 ide

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 with the same angle of projection $	heta$,the term $\frac{g}{u^2}$ must be constant.Therefore,$\frac{g_{earth}}{u_{earth}^2} = \frac{g_{planet}}{u_{planet}^2}$.Given $g_{earth} = 9.8 \, m s^{-2}$,$u_{earth} = 5 \, m s^{-1}$,and $u_{planet} = 3 \, m s^{-1}$.Substituting the values: $\frac{9.8}{5^2} = \frac{g_{planet}}{3^2}$.$\frac{9.8}{25} = \frac{g_{planet}}{9}$.$g_{planet} = \frac{9.8 	imes 9}{25} = \frac{88.2}{25} = 3.528 \, m s^{-2}$.Rounding to the nearest provided option,the value is $3.5 \, m s^{-2}$.

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