A stone is thrown straight up on Pluto, where the gravitational acceleration is 0.063g and air resistance
nonexistent. The stone returns to its starting height after 13 seconds.
a) Measured from the starting point, how high did the stone go?

To find the maximum height reached by the stone, we can use the kinematic equations. Since the stone returns to its starting height after 13 seconds, the time to reach the highest point is half of this duration, or 6.5 seconds.

Given dataTo find the maximum height reached by the stone, we can use the kinematic equations. Since the stone returns to its starting height after 13 seconds, the time to reach the highest point is half of this duration, or 6.5 seconds.

Given 
- Total time (up and down): \( t_{\text{total}} = 13 \, \text{s} \)
- Time to reach max height: \( t = \frac{13}{2} = 6.5 \, \text{s} \)
- Gravitational acceleration on Pluto: \( g_{\text{Pluto}} = 0.063 \times g_{\text{Earth}} \)

Where:
- \( g_{\text{Earth}} \approx 9.8 \, \text{m/s}^2 \), so:
  \[
  g_{\text{Pluto}} = 0.063 \times 9.8 \approx 0.6174 \, \text{m/s}^2
  \]

### Step 1: Finding Initial Velocity (\( v_0 \))
At the highest point, the final velocity \( v \) will be 0 (since the stone stops momentarily before descending). Using the kinematic equation:
\[
v = v_0 - g_{\text{Pluto}} \times t
\]
Setting \( v = 0 \):
\[
0 = v_0 - (0.6174) \times (6.5)
\]
\[
v_0 = 0.6174 \times 6.5 \approx 4.0131 \, \text{m/s}
\]

### Step 2: Finding Maximum Height (\( h \))
Now, we can find the maximum height using the kinematic formula:
\[
h = v_0 \times t - \frac{1}{2} g_{\text{Pluto}} \times t^2
\]
Substituting the values:
\[
h = (4.0131) \times (6.5) - \frac{1}{2} \times (0.6174) \times (6.5)^2
\]

Calculating each term:
1. \( 4.0131 \times 6.5 \approx 26.08515 \)
2. \( \frac{1}{2} \times 0.6174 \times 42.25 \approx 13.0437 \)

So:
\[
h = 26.08515 - 13.0437 \approx 13.04 \, \text{m}
\]

### Answer:
The maximum height reached by the stone is approximately **13.04 meters**.

Would you like more details or have any questions?

---

Here are 5 related questions for further exploration:
1. What would the maximum height be if the gravitational acceleration was doubled on Pluto?
2. How would the initial velocity change if the total time for the trip was 20 seconds instead?
3. How would the height be affected if there was air resistance on Pluto?
4. If the stone was thrown with double the initial velocity, what would the new maximum height be?
5. How much time would the stone spend in the air on Earth under similar conditions?

**Tip:** When dealing with objects in free fall, remember that the time to reach the highest point is half the total time in a round trip.

A stone is thrown straight up on Pluto, where the gravitational acceleration is 0.063g and air resistance is nonexistent. The stone returns to its starting height after 13 seconds. a) Measured from the starting point, how high did the stone go?

Discover how to calculate the maximum height of a stone thrown on Pluto, using kinematic equations and gravitational acceleration (0.063g). This problem provides a step-by-step solution involving key concepts in projectile motion and kinematics. Ideal for high school physics students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation