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Statement-1: If there were no gravitational forec, the path of the projected body always be a straight linc. Statement-2: Gravitational force makes the path of projected body always parabolic.

Short Answer

Expert verified
Both statements are correct: without gravity, path is straight; with gravity, path is parabolic.

Step by step solution

01

Understanding Statement 1

Statement 1 claims that without gravitational force, the path of a projected body is a straight line. In physics, when no external forces act on an object, it follows Newton's first law of motion and continues in a straight line with constant velocity. Therefore, without gravity, a thrown object would indeed move in a straight line.
02

Understanding Statement 2

Statement 2 claims that gravitational force results in a parabolic path for a projected body. When an object is projected into the air, gravity acts as a constant force downward, accelerating the object towards the Earth. This combination of horizontal motion and vertical acceleration due to gravity forms a parabolic trajectory.
03

Evaluating the Correctness of the Statements

Both statements relate cause (gravitational force) and effect (path shape). In the absence of gravitational force, an object would indeed move in a straight line (Statement 1 is correct). With gravitational force, the object's path follows a parabola (Statement 2 is also correct). Both statements accurately describe the motion of projectiles under specific conditions.

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Key Concepts

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

Newton's First Law of Motion
Newton's First Law of Motion, often referred to as the law of inertia, is a fundamental principle in physics. It states that an object will remain at rest or move in a straight line at a constant speed unless acted upon by a net external force.

This means that if you throw a ball in space where no external forces like gravity or air resistance exist, it will keep moving in the same direction at the same speed forever. This is because there is nothing to change its state of motion.

In our everyday lives, we rarely observe this because forces like friction and gravity are almost always at play, affecting the motion of objects. Understanding the law of inertia helps explain why a projectile would travel in a straight line if we could remove these forces, as illustrated in the first statement of the original exercise.
Gravitational Force
Gravitational force is the attraction between two masses. On Earth, this force pulls objects towards the center of the planet. It is the reason why things fall when you drop them.

Isaac Newton was the first to accurately describe the gravitational force in his law of universal gravitation. This force is crucial for understanding projectile motion. When you throw an object into the air, gravity continuously pulls it downward.
  • Gravity is constant and acts in a vertical direction, pulling projectiles down at 9.8 m/s² near the Earth's surface.
  • This causes the object to experience a downward acceleration, altering its path from straight to curved.

Without gravity, any object you threw would continue moving in the path you launched it, as a demonstration of Newton's first law. However, gravitational force curves its path, turning a straight line into a curve.
Parabolic Trajectory
A parabolic trajectory is the curved path that a projectile follows under the influence of constant gravity, assuming no air resistance.

When you throw a ball, two types of motion occur simultaneously: horizontal motion and vertical motion.

The horizontal motion is constant because no horizontal forces act on it (ignoring air resistance). The vertical motion is changed by the gravitational force, which pulls the object downwards.

This unique combination of motions results in a path that looks like a parabola. An important characteristic of this motion is that at the peak of the trajectory, the vertical velocity is zero, but the horizontal velocity remains unchanged. As a result, understanding parabolic trajectories helps predict where a projectile will land.
  • The path is symmetrical around the highest point.
  • The time to reach the peak is the same as the time to descend from it.

This concept is not only essential in physics but also in real-world applications like sports, video games, and even space exploration.

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Most popular questions from this chapter

\(\Lambda\) particle moves in \(x-y\) plane along the curve \(y=x^{3}-9 x^{2} .\) Write possible unit vector in the direction of motion, when the particle is at any point \(P(x, y)\).

In uniform circular motion a body moves with constant speed \(v\) on a circular path of constant radius \(r\). In this motion Column-I (a) The acceleration of body is (b) The kinetic energy (c) The angular displacement of body at any instant is directed along (d) The velocity of body is always directed along Column-II (p) tangent to path at evety point (q) alng axis of rotation (r) constant in magnitude but changing in direction (s) constant always

\(\Lambda\) boat is sent across (perpendicular) a river with a velocity of \(8 \mathrm{~km} / \mathrm{h}\). If the resultant velocity of the boat is \(10 \mathrm{~km} / \mathrm{h}\), the river is flowing with a velocity (a) \(6 \mathrm{~km} / \mathrm{h}\) (b) \(8 \mathrm{~km} / \mathrm{h}\) (c) \(10 \mathrm{~km} / \mathrm{h}\) (d) \(128 \mathrm{~km} / \mathrm{h}\)

'Ihe equations of motion of a body are given by \(x=36 t\) and \(2 y=96 t-9.8 t^{2}\). (lake \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\) ) (a) 'lhe height of the tower, over which it can just pass (when it is \(100 \mathrm{~m}\) from the point of projection), is \(110.6 \mathrm{~m}\). (b) 'lhe angle of projection is \(\sin ^{-1}(4 / 5)\). (c) 'lhe velocity of projection is \(60 \mathrm{~m} / \mathrm{sec}\). (d) 'lhe range on the horizontal plane is \(200 \mathrm{~m}\)

\Lambda point on the rim of a circular disc has a linear speed of \(10 \mathrm{~m} / \mathrm{s}\) at an instant when it is decreasing at the rate of \(60 \mathrm{~m} / \mathrm{s}^{2}\). If the magnitude of the total acceleration of the point at this instant is \(100 \mathrm{~m} / \mathrm{s}^{2}\), the radius of the disc is (a) \(1.25 \mathrm{~m}\) (b) \(12.5 \mathrm{~m}\) (c) \(25 \mathrm{~m}\) (d) \(2.5 \mathrm{~m}\)

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