Question

A projectile is fired from the surface of the earth with a velocity of 5 ms1 and angle with the horizontal. Another projectile fired from another planet with a velocity of 3 ms1 at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth

Answer 1

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sonuvuce Ace

Answer:

9.528 m/s²

Explanation:

Since both the projectiles follow the same path and same trajectories, their maximum height and range will be same.

Let the acceleration due to gravity on the planet be g'

Let u = 5 m/s and u' = 3 m/s

Equating the ranges of both the projectiles we get

\frac{u^{2}sin2\theta }{g} =\frac{u'^{2}sin2\theta }{g'}

or, g'=g\frac{u'^{2} }{u^{2} }

or, g' = 9.8 × \frac{3^{2} }{5^{2} }

or, g' = 9.8 × \frac{9}{25}

⇒ g' = 3.528 m/s²