The acceleration due to gravity on the moon is about one sixth its value on earth. If a baseball reaches a height of 62 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
EASd
To determine the height the baseball would reach when thrown on the surface of the moon, we can use the following formula:
H = (v^2 - u^2) / (2g)
Where:
H is the height,
v is the final velocity,
u is the initial velocity, and
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 need to find the equivalent 'g' value for the moon.
On Earth: g_earth = 9.8 m/s^2
On Moon: g_moon = 1/6 * g_earth = 1/6 * 9.8 m/s^2 ≈ 1.63 m/s^2
Now, we know that when the baseball is thrown on Earth, it reaches a height of 62 m. This implies that at the highest point, the final velocity is zero (v = 0) and the initial velocity is unknown (u).
Plugging these values into the formula, we have:
62 = (0 - u^2) / (2 * 9.8)
Simplifying the equation:
2 * 9.8 * 62 = - u^2
u^2 = - 2 * 9.8 * 62
u ≈ √(2 * 9.8 * 62)
Once we find the initial velocity on Earth, we can use it to find the height the baseball would reach on the moon.
Now, on the moon, we can use the same formula:
H_moon = (v^2 - u^2) / (2 * g_moon)
Plugging in the values:
H_moon = (0 - u^2) / (2 * 1.63)
Simplifying the equation:
H_moon ≈ - u^2 / 3.26
Finally, substituting the value of 'u' we calculated earlier:
H_moon ≈ - (√(2 * 9.8 * 62))^2 / 3.26
Evaluating the expression:
H_moon ≈ - (2 * 9.8 * 62) / 3.26
H_moon ≈ - 1207.6 / 3.26
H_moon ≈ - 370.55 m
Since height cannot be negative, we take its positive value:
The baseball would reach a height of approximately 370.55 meters when thrown on the surface of the moon.
To find the height the baseball would reach when thrown on the surface of the moon, we can use the concept of projectile motion and the fact that the acceleration due to gravity on the moon is one sixth of its value on Earth.
Let's first calculate the time it takes for the baseball to reach its maximum height on Earth. We can use the kinematic equation:
y = (v0y * t) - (0.5 * g * t^2)
where:
y = vertical displacement (height)
v0y = vertical component of the baseball's initial velocity
t = time of flight
g = acceleration due to gravity on Earth
In this case, the baseball reaches a height of 62 m on Earth. Since it is thrown upward, the final velocity at the maximum height is 0 m/s. Therefore, v0y is the vertical component of the initial velocity. We need to find this value.
Now, we know that the acceleration due to gravity on the moon is one sixth of that on Earth. So, the acceleration due to gravity on the moon (g_moon) can be calculated by multiplying the acceleration due to gravity on Earth (g_earth) by 1/6:
g_moon = (1/6) * g_earth
Next, we can calculate the time it takes for the baseball to reach its maximum height on the moon using the same kinematic equation as before, but with the acceleration due to gravity replaced by g_moon:
y_moon = (v0y_moon * t_moon) - (0.5 * g_moon * t_moon^2)
where:
y_moon = vertical displacement (height) on the moon
v0y_moon = vertical component of the baseball's initial velocity on the moon
t_moon = time of flight on the moon
Since we know that the initial and final velocities on the moon are the same as on Earth, and only the acceleration due to gravity changes, we can write:
v0y_moon = v0y_earth
t_moon = t_earth
With these equations, we can find the height the baseball would reach when thrown on the surface of the moon. Let's solve the equations:
y_moon = y * (g_moon / g_earth)
Substituting the given values:
y_moon = 62 * (1/6)
Calculating:
y_moon = 10.333... m
Therefore, the baseball would reach a height of approximately 10.33 m when thrown on the surface of the moon.