Suppose that on earth you can throw a ball vertically upward a distance of 2.00 m. Given that the acceleration of gravity on the Moon is 1.67 m/s2, how high could you throw a ball on the Moon? (Take the y-axis in the vertical direction, and assume that the location of your hand is at y = 0.)
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First, we must find the initial velocity that the ball must have on earth, to reach the height of 2 m. The height is calculated using the following equation:
H(earth) = (v^2 – v0^2)/-2g (earth)
v → is the velocity for each instant.
v0 → is the initial velocity of the ball.
g(earth) → is the gravity of the earth.
H(earth) → is the height reached on earth.
The negative sign in the denominator means that we consider the upward movement. The final velocity v at the point of maximum height is zero, so that:
H(earth) = (– v0^2)/-2g (earth)
v0 = sqrt (2*g(earth)*H(earth) )
v0 = sqrt (2*9.8*2) = 6.26 m/s
With this value of the initial velocity, we evaluated the same equation above to know the height achieved when the body is thrown on the moon.
H(moon) = (v^2 – v0^2)/-2g(moon)
H(moon) = (– v0^2)/-2g(moon) = (6.26)^2/(2*1.67) = 11.73 m
So, this ball thrown on the moon, will reach a height of 11.73 m.
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