A projectile is fired from the surface of the earth with a velocity of $5 m/s$ at an angle $θ$ with the horizontal. Another projectile, fired from another planet with a velocity of $3 m/s$ at the same angle, follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of acceleration due to gravity on the planet (in $m s^{- 2}$ ) is $({Given g = 9.8 ms}^{- 2})$

3.5

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5.9

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16.3

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110.8

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Solution

The correct option is A $3.5$
Lets assume the value of acceleration due to gravity on the planet is $g_{p}$ .

$y = x tan θ - \frac{g x^{2}}{2 u^{2} {cos}^{2} θ}$

For the trajectory of both the projectiles to be same, the ratio $\frac{g}{u^{2}}$ should be equal, as every other variable is same.

$∴ \frac{g}{u_{e}^{2}} = \frac{g_{p}}{u_{p}^{2}}$
$\frac{9.8}{5^{2}} = \frac{g_{p}}{3^{2}}$

$g_{p} = 3.5 m / s^{2}$

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