The acceleration due to gravity on the moon
is about one-sixth its value on earth.
If a baseball reaches a height of 64 m when
thrown upward by someone on the earth,
what height would it reach when thrown in
the same way on the surface of the moon?
Answer in units of m
Can you explain how you do it?
h = (6/1) * 64 = 384 m.
To find the height the baseball would reach when thrown on the moon, we can use the concept of projectile motion. The basic formula for the maximum height attained by a projectile is given by:
H = (vₒ² * sin²θ) / (2g)
Where:
H is the maximum height
vₒ is the initial velocity of the projectile
θ is the launch angle
g is the acceleration due to gravity
Given that the acceleration due to gravity on the moon is one-sixth its value on Earth, we can substitute 1/6g for g in the formula.
Now, we are given that the baseball reaches a height of 64 m when thrown on Earth. However, we do not have the launch angle or the initial velocity of the baseball. Without this information, we cannot directly calculate the height it would reach on the moon.
If we assume that the baseball is thrown with the same launch angle and initial velocity on both Earth and the moon, we can determine the new height by using the proportional relationship between the acceleration due to gravity on Earth and the moon.
Since g on the moon is one-sixth of that on Earth, the maximum height on the moon would be six times greater. Therefore, on the moon, the baseball would reach a height of:
H(moon) = H(earth) × 6
Substituting the given value, we have:
H(moon) = 64 m × 6
H(moon) = 384 m
So, when thrown in the same way on the surface of the moon, the baseball would reach a height of 384 meters.