a projectile is fired from the surface of the earth with a velocity of 5 metre per second and angle theta with the horizontal another projectile fired from another planet with a velocity of 3 metre per second 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
Answers
Answer:
3.6 m/s²
Explanation:
a projectile is fired from the surface of the earth with a velocity of 5 metre per second and angle theta with the horizontal another projectile fired from another planet with a velocity of 3 metre per second 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
At Earth
Horizontal Velocity = 5Cosθ
Vertical Velocity = 5Sinθ
a = g = 10 m/s²
Time to reach Peak
0 = 5Sinθ - 10T
=> T = Sinθ/2
Horizontal Distance at Peak = 5Cosθ*Sinθ/2
Vertical Distance at Peak = 5SinθSinθ/2 = 5Sin²θ/2
At another Planet
Horizontal Velocity = 3Cosθ
Vertical Velocity = 3Sinθ
a m/s²
Time to reach Peak
0 = 3Sinθ - aT
=> T = 3Sinθ/a
Horizontal Distance at Peak = 3Cosθ*3Sinθ/a = 9Cosθ*Sinθ/a
Vertical Distance at Peak = 3Sinθ3Sinθ/a = 9Sin²θ/a
as Trajectories are same so Horizontal & Vertical distance at peak are equal
=> 5Cosθ*Sinθ/2 = 9Cosθ*Sinθ/a
=> a = 18/5 = 3.6
or
5Sin²θ/2 = 9Sin²θ/a => a = 3.6
acceleration due to gravity on the planet is 3.6 m/s²
Explanation:
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