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Chapter 4: Problem 48

A body is thrown upwards from the earth surface with velocity \(5 \mathrm{~ms}^{-1}\) and from a planet surface with velocity \(3 \mathrm{~ms}^{-1}\). Both follow the same path. What is the projectile acceleration due to gravity on the planet? Acceleration due to gravity on earth is \(10 \mathrm{~ms}^{-1} .\) [Orissa JEE 2008] (a) \(2 \mathrm{~ms}^{-2}\) (b) \(3.6 \mathrm{~ms}^{-2}\) (c) \(4 \mathrm{~ms}^{-2}\) (d) \(5 \mathrm{~ms}^{-2}\)

Short Answer

Expert verified
The projectile acceleration due to gravity on the planet is \(3.6 \text{ m/s}^2\) (Option b).

Step by step solution

01

Identify Given Information and Variables

The initial velocity on Earth is given as \( v_{e} = 5 \text{ m/s} \) and on the planet as \( v_{p} = 3 \text{ m/s} \). The acceleration due to gravity on Earth is \( g_{e} = 10 \text{ m/s}^2 \). We need to find the acceleration due to gravity on the planet \( g_{p} \).
02

Understand the Problem

The problem states that both bodies follow the same path. This implies that they reach the same maximum height. We can use the formula for maximum height in projectile motion: \( h = \frac{v^2}{2g} \).
03

Set Up the Equation for Maximum Height

The height reached on Earth can be expressed as \[ h = \frac{v_{e}^2}{2g_{e}} = \frac{5^2}{2 \times 10} = \frac{25}{20} = 1.25 \text{ meters} \]. The height reached on the planet is \[ h = \frac{v_{p}^2}{2g_{p}} = \frac{3^2}{2g_{p}} = \frac{9}{2g_{p}} \]. Since both heights are equal, set the equations equal: \[ \frac{9}{2g_{p}} = 1.25 \].
04

Solve for Planet's Gravity

Rearrange and solve the equation: \[ 2g_{p} = \frac{9}{1.25} \]. This simplifies to \( 2g_{p} = 7.2 \). Finally, solving for \( g_{p} \), we get \[ g_{p} = \frac{7.2}{2} = 3.6 \text{ m/s}^2 \].

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Key Concepts

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

Acceleration due to Gravity
The acceleration due to gravity, often denoted by the letter "g," is a crucial concept in understanding how objects move under the influence of Earth's or any other celestial body's gravitational pull. On Earth, this value is approximately
  • 10 meters per second squared (\(10 \text{ m/s}^2\)).
  • This means that for every second an object is in free fall, its velocity increases by 10 meters per second, provided there's no air resistance.
On a different planet, the acceleration due to gravity can be more or less than on Earth, influencing the motion of projectiles. When solving physics problems involving projectile motion, knowing the acceleration due to gravity on the respective planet is essential as it affects the time and distance that the projectile will travel.
In our original exercise, calculating the acceleration due to gravity on a different planet required comparing the maximum height reached by a projectile on Earth and on the planet, given their initial velocities.
Maximum Height in Projectile Motion
In projectile motion, determining the maximum height is important for understanding the trajectory or path of an object thrown into the air. The maximum height (\( h \)) can be calculated using the formula:
  • \( h = \frac{v^2}{2g} \)
where:
  • \( v \) is the initial velocity of the projectile,
  • \( g \) is the acceleration due to gravity.
By rearranging this formula, you can calculate the height an object reaches given its initial speed and gravitational acceleration.
In the exercise scenario, the maximum height reached by the projectile was used to compare motion on Earth and a planet. Since both projectiles reached the same height, we equated the heights derived from their respective initial velocities and gravitational accelerations, allowing us to solve for the unknown gravity on the planet.
Kinematic Equations
Kinematic equations are mathematical relationships that describe the motion of objects without considering forces causing the motion. They are widely used in solving problems related to projectile motion, free-falling objects, and more. The kinematic equation for vertical motion involving maximum height is demonstrated as:
  • \( h = \frac{v^2}{2g} \)
This specific equation helps in finding how high a projectile reaches, given its initial speed and the acceleration due to gravity acting on it.
Using kinematic equations is helpful to derive parameters about an object's motion, such as the maximum height, time of flight, and range, under conditions of constant acceleration. They provide a robust framework for solving complex problems by offering simplified, focused calculations.
In our problem, the kinematic equation helped establish a relationship between the initial speeds on Earth and the planet, leading us to find the unknown gravitational acceleration on the planet.

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Most popular questions from this chapter

A stone is projected with a velocity \(20 \sqrt{2} \mathrm{~ms}^{-1}\) at an angle of \(45^{\circ}\) to the horizontal. The average velocity of stone during its motion from starting point to its maximum height is \(\left(g=10 \mathrm{~ms}^{-2}\right)\) (a) \(5 \sqrt{5} \mathrm{~ms}^{-1}\) (b) \(10 \sqrt{5} \mathrm{~ms}^{-1}\) (c) \(20 \mathrm{~ms}^{-1}\) (d) \(20 \sqrt{5} \mathrm{~ms}^{-1}\)

The maximum height attained by projectile is (a) \(2 \mathrm{~h} / 3\) (b) \(3 h\) (c) \(3 h / 4\) (d) \(3 h / 2\) Two second after projection, a projectile is travelling in a direction inclined at \(30^{\circ}\) to the horizontal. After I more second, it is travelling horizontally (use \(g=10 \mathrm{~ms}^{-2}\) )

Assertion-Reason type. Each of these contains two Statements: Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choices, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given ahead (a) If both Assertion and Reason are true and Reason is correct explanation of the Assertion (b) If both Assertion and Reason are true but Reason is not correct explanation of the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true Assertion A particle is projected with speed \(u\) at an angle \(\theta\) with the horizontal. At any time during motion, speed of particle is \(v\) at angle \(\alpha\) with the vertical, then \(v \sin \alpha\) is always constant throughout the motion. Reason In case of projectile motion, magnitude of radial acceleration at top most point is maximum.

The kinetic energy of a project at the height point is half of the initial kinetic energy. What is the angle of projection with the horizontal? [a) \(3 \overline{0^{\circ}}\) (b) \(45^{*}\) (c) \(6 \overline{0^{n}}\) (d) \(90^{*}\)

When a projectile is projected at a certain angle with the horizontal, its horizontal range is \(R\) and time of flight is \(T_{1}\). When the same projectile is throwing with the same speed at some other angle with the horizontal, its horizontal range is \(R\) and time of flight is \(T_{2}\). The produet of \(T_{1}\) and \(T_{2}\) is (a) \(\frac{R}{g}\) (b) \(\frac{2 R}{g}\) (c) \(\frac{3 R}{g}\) (d) \(\frac{4 R}{g}\)
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