The acceleration due to gravity on the moon

is about one-sixth its value on earth.
If a baseball reaches a height of 78 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

h = 78 * 6 = 468 m.

To find the height a baseball would reach when thrown on the surface of the moon, we need to consider the difference in acceleration due to gravity between the moon and the Earth.

Given that the acceleration due to gravity on the moon is about one-sixth its value on Earth, we can calculate the acceleration due to gravity on the moon (g_moon) using the equation:

g_moon = (1/6) * g_earth

Since the baseball reaches a height of 78 m when thrown on Earth, we can use the kinematic equation for vertical motion to find the time it takes for the baseball to reach its peak height:

v_final = v_initial + (g_earth * t)

At the peak height, the final velocity (v_final) will be 0 m/s since the baseball momentarily stops before falling back down. Therefore, we can rearrange the equation to solve for the time (t):

t = -v_initial / g_earth

Now, we can use the time (t) to calculate the height (h_moon) the baseball would reach when thrown on the moon using the following equation:

h_moon = (1/2) * g_moon * t^2

Let's substitute the values and calculate the result:

g_earth = 9.8 m/s^2 (acceleration due to gravity on Earth)
h_earth = 78 m (height reached when thrown on Earth)

g_moon = (1/6) * 9.8 m/s^2 (acceleration due to gravity on the moon)
t = -v_initial / g_earth (time taken to reach peak height on Earth)
h_moon = (1/2) * g_moon * t^2 (height reached when thrown on the moon)

First, let's calculate the time it takes for the baseball to reach its peak height on Earth:

t = -v_initial / g_earth

v_initial = 0 m/s (since it is thrown upward initially)

t = -0 / 9.8
t = 0

Since the time calculated is zero, it means that the baseball reaches its peak height instantaneously without any time duration.

Now, let's calculate the height the baseball would reach when thrown on the moon:

h_moon = (1/2) * g_moon * t^2

h_moon = (1/2) * (1/6) * 9.8 * 0^2
h_moon = 0 m

Therefore, the height the baseball would reach when thrown on the surface of the moon is 0 meters.