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In ideal projectile motion, when the positive \(y\) -axis is chosen to be vertically upward, the \(y\) -component of the velocity of the object during the ascending part of the motion and the \(y\) -component of the velocity during the descending part of the motion are, respectively, a) positive, negative. b) negative, positive. c) positive, positive. d) negative, negative.

Short Answer

Expert verified
In an ideal projectile motion, the y-component of velocity during the ascending part is positive, representing the upward motion of the projectile against the force of gravity. As the object moves upward, gravity causes the y-component of velocity to decrease until it reaches zero at the peak of the trajectory. During the descending part, the y-component of velocity is negative, indicating that the projectile is accelerating downward due to the force of gravity. The understanding of these velocity components allows us to analyze and predict the behavior of projectiles in motion.

Step by step solution

01

Projectile motion basic theory

An ideal projectile motion assumes no air resistance or any other force besides gravity acting on the projectile. Total motion will be a parabola under these conditions. The horizontal velocity remains constant, while the vertical velocity is subjected to gravity and will change during the projectile's motion.
02

Initial Velocity Components

as we have \(v_0\) as the initial velocity, we can find the x-component of the velocity (\(v_{0x}\)) and the y-component of the velocity (\(v_{0y}\)) from the initial velocity angle \(\alpha\). \(v_{0x} = v_0 \cos{\alpha}\), \(v_{0y} = v_0 \sin{\alpha}\).
03

Ascending Part - y-component of velocity

During the ascending part of the projectile motion, the object moves against the force of gravity. The gravitational force will cause a vertical component of the acceleration, and this acceleration will be negative (\(-g\)) towards the Earth. The y-component of velocity decreases as the object ascends, and reaches to 0 at the peak of its trajectory. Since the object is moving upwards, the \(y\)-component of the velocity during the ascending part is positive.
04

Descending Part - y-component of velocity

As the object starts its descent, the force of gravity accelerates the object downwards, the y-component of velocity increases in the direction of the Earth. The gravitational force is acting in the same direction as the motion, so the y-component of the velocity during the descending part is negative.
05

Conclusion

Based on the analysis of ascending and descending parts of an ideal projectile motion, the correct answer based on the given options is: a) The y-component of the velocity during the ascending part of the motion is positive, and the y-component of the velocity during the descending part of the motion is negative.

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Key Concepts

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

Understanding Initial Velocity Components
When unpacking the world of projectile motion, the concept of 'initial velocity components' is crucial. Let's take a ball thrown at an angle into the air as an example. The initial velocity is the speed it leaves your hand, and this speed can be divided into two parts or 'components'. The horizontal component (\(v_{0x}\)) and the vertical component (\(v_{0y}\)). These components are calculated from the initial throw angle (\( \theta \theta\text{ using simple trigonometry: }

\(v_{0x} = v_0 \cos(\theta)\),
\(v_{0y} = v_0 \sin(\theta)\).

This initial push is what sets the projectile on its course. Understanding these components helps us predict how far and how high the object will travel. The horizontal component remains unchanged throughout the flight (ignoring air resistance), while the vertical component is deeply affected by gravity, leading us to the next fundamental concept.
Vertical Acceleration due to Gravity
Gravity is what makes an apple fall from a tree, and it's the same force that acts on our hypothetical ball in projectile motion. This force gives the object a constant vertical acceleration towards Earth, which we represent as '\(g\)'. Interesting to know, \(g\) is approximately \(9.81 \, m/s^2\) going downward, no matter the initial throw.

Now, what does this mean for our object in motion? It makes the upward journey a battle against gravity, where the vertical speed decreases until it reaches the peak and becomes zero for a brief moment. From there, gravity takes over fully, pulling the object back down, increasing its vertical speed until it either lands or encounters an interruption. This acceleration does all sorts of things to the velocity, giving us a way to calculate both height and time in the air. Always remember—an object's vertical speed changes due to gravity, but its horizontal speed remains unaffected in ideal conditions.
Ascending and Descending Velocity
Breaking down the ascent and descent can clarify a lot about projectile motion. During the ascent, or the 'going up' phase, the velocity component is in opposition to gravity. Think of it as a slowing down vehicle that eventually stops for a split second before descending. In mathematics, this slowing motion is represented as a decreasing positive value.

When the projectile reaches its highest point, the vertical velocity, for an instant, hits zero. It's the turning point where the descent begins. Gravity doesn't snooze - the object accelerates downwards, picking up speed. This time, our math shows an increasing negative value for that vertical component of velocity. It's a neat symmetrical fashion: what gains in speed during the fall mirrors what it loses going up. So, always imagine that symmetrical parabolic path it traces against the sky—an arcing journey up and back down, governed by the immutable law of gravity.

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Most popular questions from this chapter

A projectile is launched from the top of a building with an initial velocity of \(30.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(60.0^{\circ}\) above the horizontal. The magnitude of its velocity at \(t=5.00 \mathrm{~s}\) after the launch is a) \(-23.0 \mathrm{~m} / \mathrm{s}\). b) \(7.3 \mathrm{~m} / \mathrm{s}\). c) \(15.0 \mathrm{~m} / \mathrm{s}\). d) \(27.5 \mathrm{~m} / \mathrm{s}\). e) \(50.4 \mathrm{~m} / \mathrm{s}\).

During the 2004 Olympic Games, a shot putter threw a shot put with a speed of \(13.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(43.0^{\circ}\) above the horizontal. She released the shot put from a height of \(2.00 \mathrm{~m}\) above the ground. a) How far did the shot put travel in the horizontal direction? b) How long was it until the shot put hit the ground?

A security guard is chasing a burglar across a rooftop, both running at \(4.20 \mathrm{~m} / \mathrm{s} .\) Before the burglar reaches the edge of the roof, he has to decide whether or not to try jumping to the roof of the next building, which is \(5.50 \mathrm{~m}\) away and \(4.00 \mathrm{~m}\) lower. If he decides to jump horizontally to get away from the guard, can he make it? Explain your answer.

In a projectile motion, the horizontal range and the maximum height attained by the projectile are equal. a) What is the launch angle? b) If everything else stays the same, how should the launch angle, \(\theta_{0},\) of the projectile be changed for the range to be halved?

The captain of a boat wants to travel directly across a river that flows due east with a speed of \(1.00 \mathrm{~m} / \mathrm{s} .\) He starts from the south bank of the river and heads toward the north bank. The boat has a speed of \(6.10 \mathrm{~m} / \mathrm{s}\) with respect to the water. In what direction (in degrees) should the captain steer the boat? Note that \(90^{\circ}\) is east, \(180^{\circ}\) is south, \(270^{\circ}\) is west, and \(360^{\circ}\) is north.

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