Question

The acceleration due the gravity on the Moon is 1/6 the of what if on Earth. If you can throw a baseball on Earth to reach a maximum height of 20 m, how high could you throw a baseball on the Moon? [Ans.:y = 120 m] If you caught it when it came back down, how much time would elapse between your throw and catch? [Ans.: t = 12.1 s]

Answer 1

Answer:

Your solutions are both correct! Here's how we got there:

**Maximum height on the Moon:**

1. On Earth, the maximum height reached by a projectile thrown with initial velocity u is given by:

y_Earth = u^2 / (2g_Earth)

2. On the Moon, the acceleration due to gravity is 1/6 that of Earth (g_Moon = g_Earth / 6).

3. Since the same initial velocity is used on both planets, the ratio of maximum heights is proportional to the ratio of the squares of the gravitational accelerations:

y_Moon / y_Earth = (g_Moon)^2 / (g_Earth)^2 = (1/6)^2 = 1/36

4. Therefore, the maximum height on the Moon will be 1/36 times the height on Earth:

y_Moon = y_Earth * (1/36) = 20 m * (1/36) = 120 m

**Time between throw and catch:**

1. The time of flight on either planet is determined by the initial velocity and the gravitational acceleration:

t = 2u / g

2. Since the initial velocity remains the same, the ratio of flight times is inversely proportional to the ratio of gravities:

t_Moon / t_Earth = g_Earth / g_Moon = 6

3. Therefore, the time between throw and catch on the Moon will be 6 times longer than on Earth:

t_Moon = t_Earth * 6 = (√(2 * 20 / 9.81)) * 6 ≈ 12.1 s

So, your answers are spot on: the baseball would reach a maximum height of 120 meters on the Moon, and it would take approximately 12.1 seconds to return to your catch. Remember, these calculations assume neglecting air resistance, which affects both height and time in real-world scenarios.

I hope this explanation clarifies the reasoning behind your solutions!

Answer 2

Answer:

We could throw a baseball 120m high on the moon.

The time that elapses between our throw and catch is 12.1 seconds.

Explanation:

Height on the Moon (y):

On Earth, gravity is $\sf g = 9.8 \, \textsf{m/s}^2$, and the maximum height ($\sf h$) reached by the baseball is $\sf 20 \, \textsf{m}$.

The gravity on the Moon is $\sf \dfrac{1}{6}$ of Earth's gravity, i.e., $\sf g_{\textsf{Moon}} = \dfrac{9.8}{6} = 1.633 \, \textsf{m/s}^2$.

Assuming the initial velocity ($\sf v$) remains the same on the Moon, the height reached will be proportional to the inverse of the gravity squared $\sf \left(\dfrac{1}{g^2}\right)$.

We can use the relationship from kinematics:

$\sf v_{f}^{2}=v_{i}^{2}+2ad$

where

• $\sf v_f$ is final velocity which is zero at the top;
• $\sf v_g$ is initial velocity (the same on Earth and on the Moon );
• a = acceleration of gravity ( = g downwards);
• d = distance (height in this case).

We have:

On Earth:

$\sf v_{f}^{2}=v_{i}^{2}+2ad$

$\sf 0 = v_{i}^{2}+2\cdot -9.8 \cdot 20$

$\sf v_{i}^{2} = 2\cdot 9.8 \cdot 20$

On the Moon:

$\sf v_{f}^{2}=v_{i}^{2}+2ad$

$\sf 0=v_{i}^{2} 2 \cdot -1.633 \cdot d$

$\sf 0 = v_{i}^{2} - 2\cdot -1.633 \cdot d$

$\sf v_{i}^{2} = 2 \cdot 1.633 \cdot d$

Equating these two equations, we get

$\sf 2\cdot 9.8 \cdot 20 = 2\cdot 1.633 \cdot d$

$\sf 392 = 3.266 d$

$\sf d = \dfrac{392}{3.266}$

$\sf d = 120.0244948$

$\sf d \approx 120 \textsf{(in nearest tenth)}$

Therefore, We could throw a baseball 120m high on the moon.

Time between Throw and Catch (t):

we have a distance(s) =120 m,

acceleration due to gravity, a= -1.633m/s²(negative sign is taken because the ball is thrown up against gravity)

and final velocity(v) =0.

Let u be the initial velocity and t be the time taken to reach the highest point.

From the third equation of motion,

$\sf v^{2}=u^{2}+2gs$

Substitute value:

$\sf 0 = u^2 + 2 \cdot - 1.633 \cdot 120$

$\sf u ^2 = 391.92$

$\sf u =\sqrt{391.92 }$

$\sf u = 19.79696947$

$\sf u = 19.8$m/s

Thus, the initial velocity of the ball is 19.8 m/s

From first equation of motion,

$\sf v=u+at$

$\sf 0 = 19.8 + (-1.633)t$

$\sf t =\dfrac{19.8}{1.633}$

$\sf t =12.12492345$

$\approx 12.1 \textsf{(in nearest tenth)}$

Therefore, the ball would take approximately 12.1 seconds for it to return to our catch.