Answer:
greater
Explanation:
Let the initial velocity is u .
Acceleration due to gravity on earth, ge = g
Acceleration due to gravity on moon, gm = g / 6
Let the maximum height reached on earth is h and the maximum height reached on the moon is h'.
Use third equation of motion
[tex]v^{2}=u^{2}- 2gh[/tex]
At maximum height, the final velocity = 0.
For earth
0 = u^2 - 2 gh
h = u^2 / 2g ..... (1)
For moon
0 = u^2 - 2 g/6 x h'
h' = 6u^2 / 2g = 6 h (From (1)
So, the maximum height reached at moon is 6 times the maximum height reached at earth.
The maximum height it reaches on the Moon is greater than the maximum height reached on the Earth.
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Acceleration is rate of change of velocity.
[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]
[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]
a = acceleration ( m/s² )
v = final velocity ( m/s )
u = initial velocity ( m/s )
t = time taken ( s )
d = distance ( m )
Let us now tackle the problem!
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We will use following formula to solve this problem:
[tex]v^2 = u^2 - 2gh[/tex]
[tex]0^2 = u^2 - 2gh[/tex]
[tex]u^2 = 2gh[/tex]
[tex]\large {\boxed {h = \frac{u^2}{2g}}}[/tex]
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From above formula , we can conclude that the maximum height is inversely proportional to the gravitational acceleration.
Gravitation acceleration on the moon is less that gravitational acceleration on the earth. Therefore , the maximum height it reaches on the Moon is greater than the maximum height reached on the Earth.
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Grade: High School
Subject: Physics
Chapter: Kinematics
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Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Projectile , Motion , Horizontal , Vertical , Release , Point , Ball , Wall
