A projectile is fired from the surface of the earth with a velocity of and angle with the horizontal. Another projectile fired from another planet with a velocity of at the same angle follows a trajectory which is identical with the trajectory of the p

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Solution

Q.
A projectile is fired from the surface of the earth with a velocity of $5 {ms}^{- 1}$ and angle $θ$ with the horizontal. Another projectile fired from another planet with a velocity of $3 {ms}^{- 1}$ at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in ${ms}^{- 2}$ ) is (Given $g = 9.8 {ms}^{- 2}$ ) [2014]

1 3.5

2 5.9

3 16.3

4 110.8

Ans.
(1)

The equation of trajectory is $y = x \tan θ - \frac{g x^{2}}{2 u^{2} \cos^{2} θ}$

where $θ$ is the angle of projection and $u$ is the velocity with which projectile is projected.

For equal trajectories and for same angles of projection,

$\frac{g}{u^{2}} = constant$

As per question, $\frac{9.8}{5^{2}} = \frac{g'}{3^{2}}$

where $g'$ is acceleration due to gravity on the planet.

$g' = \frac{9.8 \times 9}{25} = 3.5 {ms}^{- 2}$

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