/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 A ball is thrown in a projectile... [FREE SOLUTION] | 91Ó°ÊÓ

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A ball is thrown in a projectile motion trajectory with an initial velocity \(v\) at an angle \(\theta\) above the ground. If the acceleration due to gravity is \(-g,\) which of the following is the correct expression of the time it takes for the ball to reach its highest point, \(y,\) from the ground? (A) \(v^{2} \sin t / \mathrm{g}\) (B) \(-v \cos \theta / g\) (C) \(v \sin \theta / g\) (D) \(v^{2} \cos \theta / g\)

Short Answer

Expert verified
The correct expression for the time it takes for the ball to reach the highest point from the ground is \(v \sin \theta / g\). Hence, the correct option is (C).

Step by step solution

01

Identifying the formula for time of flight in vertical direction

In the vertical direction, the ball is projected with an initial velocity component of \(v \sin \theta\) upwards, and undergoes uniform acceleration downwards due to gravity (\(g\)). Hence, the formula for time of flight to reach the highest point in vertical direction is given by \(t = \frac{Initial Velocity}{Acceleration}\)
02

Substituting the values

We substitute the values into the formula from the first step. In this case, the initial velocity in vertical direction is \(v \sin \theta\) and the acceleration due to gravity is \(g\). So we find that \(t = \frac{v \sin\theta}{g}\).
03

Comparing with the given options

On comparing the derived expression with the provided choices, it is evident that option (C) matches the derived expression perfectly and hence is the correct choice.

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Key Concepts

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

Initial Velocity
When we talk about projectile motion, the term 'initial velocity' plays a crucial role in determining the path of the projectile. The initial velocity, often represented as \( v \), is the speed at which an object starts its motion. It has two components: a horizontal component \( v \cos \theta \) and a vertical component \( v \sin \theta \).
For the context of vertical motion, it is the vertical component \( v \sin \theta \) that is particularly important. This component determines how high and how long the projectile will stay in the air. It acts against gravity, propelling the object upward initially.
In problems involving projectile motion, identifying the initial velocity and breaking it into its components is the first step in analyzing the motion.
Vertical Motion
Vertical motion in projectile physics refers to the motion of a projectile in the vertical direction under the influence of gravity. This direction is usually perpendicular to the horizontal plane, such as the ground.
When a projectile is launched, its vertical motion is characterized by an upward climb followed by a descent, thanks to gravity. Initially, the projectile has an upward velocity component \( v \sin \theta \). As it moves upward, gravity, which acts in the opposite direction, slows it down until it reaches a point where the velocity becomes zero. This point is the projectile's highest point or apex.
The time and distance a projectile covers vertically are affected by its initial velocity and the acceleration due to gravity. In calculating vertical motion, one must consider both the initial velocity and gravity to understand how fast the object decelerates.
Time of Flight
The time of flight for the highest point in vertical motion is a key concept in projectile problems. It refers to the total time taken by the projectile to reach its peak height from the launch point.
To calculate this, we use the equation involving the initial vertical velocity and gravity: \( t = \frac{v\sin\theta}{g} \). Here, \( t \) is the time of flight to the highest point, \( v\sin\theta \) represents the initial vertical velocity, and \( g \) is the acceleration due to gravity.
This equation highlights that the time to the highest point depends on how powerful the initial vertical motion is relative to the gravitational pull. A larger initial vertical velocity or a smaller gravitational acceleration would result in a longer time to reach the highest point.
Acceleration due to Gravity
The term "acceleration due to gravity" is abbreviated as \( g \), and it typically takes the value of \( 9.8 \text{ m/s}^2 \) on Earth. Gravity is a constant force that pulls objects towards the center of the Earth, affecting how all objects move vertically.
In the context of projectile motion, gravity decelerates the projectile's upward motion until it stops momentarily at the peak before accelerating it back downwards. This constant force ensures that all vertical motion follows a symmetrical path, meaning the ascent to the apex mirrors the descent back to the launch height.
Understanding this concept allows us to predict and calculate various aspects of projectile motion, such as how long a projectile will stay aloft and how its upward movement balances out with the downward pull of gravity.

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