First slide

Projection Under uniform Acceleration

Question

A projectile is fired from the surface of the earth with a velocity of $5 m s^{- 1}$ and angle $θ$ with the horizontal. Another projectile fired from another planet with a velocity of $3 m s^{- 1}$ 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 $(m s^{- 2})$ is (Given g = $m s^{- 2}$

Moderate

A

3.5
B

5.9
C

16.3
D

110.8

Solution

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 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 the gravity on the planet.
$g^{'} = \frac{9.8 \times 9}{25} = 3.5 {ms}^{- 2}$

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