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Chapter 4: Problem 42

An astronaut is on the surface of a planet whose air resistance is negligible. To measure the acceleration due to gravity \((\mathrm{g})\), he throws a stone upwards. He observer that the stone reaches to a maximum height of \(\mathrm{h}=10 \mathrm{~m}\) (which is negligible is compare radius of planet) and reaches the surface 4 second after it was thrown. Find the acceleration due to gravity (g) on the surface of that planet : (1) \(5 \mathrm{~m} / \mathrm{s}^{2}\) (2) \(10 \mathrm{~m} / \mathrm{s}^{2}\) (3) \(7.5 \mathrm{~m} / \mathrm{s}^{2}\) (4) \(2.5 \mathrm{~m} / \mathrm{s}^{2}\)

Short Answer

Expert verified
The acceleration due to gravity on the planet's surface is 5 m/s².

Step by step solution

01

- Understanding the motion

The stone is thrown upwards and reaches the maximum height of 10 meters before coming back to the surface. The total time of flight is 4 seconds. Since the motion can be divided into two equal parts (ascent and descent), the time to reach the maximum height is 2 seconds.
02

- Use kinematic equation for ascent

For the stone's ascent from the surface to the maximum height, use the kinematic equation:
03

- Use kinematic equation for ascent

Final velocity v=0 at the maximum height, initial velocity u=v_0, and acceleration a=-g. We have

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

kinematic equations
In physics, kinematic equations describe the motion of objects. These equations help us relate properties such as displacement, initial velocity, final velocity, acceleration, and time. There are four primary kinematic equations, and they assume constant acceleration. In our problem, we use these equations to analyze the motion of a stone thrown upwards by an astronaut. Here are some key kinematic equations:

1. \[ v = u + at \]
2. \[ s = ut + \frac{1}{2}at^2 \]
3. \[ v^2 = u^2 + 2as \]
4. \[ s = vt - \frac{1}{2}at^2 \]
In these equations:
  • \

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