The acceleration due to gravity on the moon
is about one-sixth its value on earth.
If a baseball reaches a height of 50 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.
the K.E. of the ball becomes P.E. at its peak
1/2 m v^2 = m g h
the only difference on the moon is that g is 1/6 of Earth
so h will be six times the Earth value
To find the height the baseball would reach when thrown on the surface of the moon, we can use the concept of projectile motion.
First, let's find the initial velocity of the baseball when thrown on the moon's surface. Since the question states that the acceleration due to gravity on the moon is one-sixth of its value on Earth, we can calculate the initial velocity using the formula:
v = u + at
Where:
v = final velocity (0 m/s at the maximum height)
u = initial velocity
a = acceleration due to gravity on the moon
t = time taken to reach the maximum height
Since the final velocity at the maximum height is 0 m/s, we can rewrite the formula as:
0 = u - (1/6)g*t
Where g is the acceleration due to gravity on Earth. Rearranging the equation, we get:
u = (1/6)g*t
Now, let's determine the time taken to reach the maximum height. Since the motion is symmetrical, the time taken to reach the maximum height is equal to the time taken to fall from the maximum height back to the starting point. We can use the equation for time:
t = 2 * (v / a)
Where:
v = final velocity (0 m/s at the maximum height)
a = acceleration due to gravity on the moon
Substituting the values, we get:
t = 2 * (0 / (1/6)g)
Simplifying, we find:
t = 0
This means that the time taken to reach the maximum height is 0 seconds, as the object doesn't experience any acceleration while moving upwards.
Now, let's find the height the baseball would reach when thrown on the moon's surface using the equation:
h = ut + (1/2)at^2
Since the time taken is 0 seconds, the equation simplifies to:
h = 0 + (1/2)(1/6)g(0)^2
Simplifying further:
h = 0
Therefore, the baseball would not reach any height when thrown on the surface of the moon. It would simply fall back to the surface without reaching any maximum height.