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A football is thrown with an initial upward velocity component of \(15.0 \mathrm{~m} / \mathrm{s}\) and a horizontal velocity component of \(18.0 \mathrm{~m} / \mathrm{s}\). (a) How much time is required for the football to reach the highest point in its trajectory? (b) How high does it get above its release point? (c) How much time after it is thrown does it take to return to its original height? How does this time compare with what you calculated in part (a)? Is your answer reasonable? (d) How far has the football traveled horizontally from its original position?

Short Answer

Expert verified
(a) 1.53 s; (b) 11.5 m; (c) 3.06 s; (d) 55.08 m.

Step by step solution

01

Understanding the Vertical Motion

To find the time to reach the highest point, we need to consider only the vertical motion because the horizontal component of the velocity does not affect the height. The vertical velocity component is given as \(v_{0y} = 15.0 \, \text{m/s}\). At the highest point, the vertical velocity is \(0 \, \text{m/s}\). We will use the equation \(v = v_{0y} - gt\), where \(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity) and solve for \(t\).
02

Calculating Time to Reach the Highest Point

Set \(v = 0\) in the equation from Step 1 and solve for \(t\): \[ 0 = 15.0 - 9.8t \] Solving for \(t\), we find: \[ t = \frac{15.0}{9.8} \approx 1.53 \, \text{s} \] This is the time required for the football to reach its highest point.
03

Calculating Maximum Height

To find how high the ball gets above the release point, use the equation for vertical displacement: \[ h = v_{0y}t - \frac{1}{2}gt^2 \] Substitute \(t = 1.53 \, \text{s}\): \[ h = 15.0 \times 1.53 - \frac{1}{2} \times 9.8 \times 1.53^2 \] Solving this gives: \[ h \approx 11.5 \, \text{m} \]
04

Calculating Total Time for Return

The total time to return to the original height is twice the time to the highest point because the motion is symmetric. Therefore: \[ t_{\text{total}} = 2 \times 1.53 \approx 3.06 \, \text{s}\]
05

Checking Time Reasonableness

The calculated total time \(3.06 \, \text{s}\) is twice the time to reach the highest point, confirming the symmetry of projectile motion. This is reasonable as the ball rises and falls symmetrically.
06

Calculating Horizontal Distance

The horizontal distance can be found using the horizontal velocity component and the total time of flight. Use the formula: \[ x = v_{0x}t_{\text{total}} \] Substitute \(v_{0x} = 18.0 \, \text{m/s}\) and \(t_{\text{total}} = 3.06 \, \text{s}\): \[ x = 18.0 \times 3.06 \approx 55.08 \, \text{m} \] This is the horizontal distance traveled.

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Key Concepts

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

Vertical Motion
Vertical motion in projectile scenarios is influenced predominantly by gravity. When a football is thrown upwards, it will slow down until it reaches its peak height, then accelerate downward. This is due to gravity acting against the upward motion. In this case, the football is thrown with an initial vertical velocity of 15.0 m/s. As it ascends, gravity gradually reduces its speed until it reaches zero at the highest point. This happens because gravity, with an acceleration of 9.8 m/s², is continuously pulling the ball downwards.

For calculating how long it takes for the football to reach its highest point, the formula used is: \[ v = v_{0y} - gt \]Where,- \( v \) is the final velocity (0 m/s at the highest point),- \( v_{0y} \) is the initial vertical velocity,- \( g \) is the acceleration due to gravity (9.8 m/s²),- \( t \) is time.Plug these into the equation and solve for \( t \). This calculation shows that the vertical motion is a direct battle between the ball's initial speed and gravity's pull.
Horizontal Velocity
Horizontal velocity in projectile motion remains unaffected by gravity. This is key to understanding projectile motion. When a football is thrown, its horizontal motion continues at a constant speed because, unlike vertical motion, there are no opposing forces like gravity acting in the horizontal direction.

Here, the football has a horizontal velocity of 18.0 m/s. This means during its entire flight, it travels horizontally at this speed. The total time the football stays in the air, determined by vertical motion, is crucial for computing how far it travels horizontally.

The horizontal distance \( x \) the ball covers is given by:\[ x = v_{0x} \, t_{\text{total}} \]where,- \( v_{0x} \) is the initial horizontal velocity,- \( t_{\text{total}} \) is the total time in the air, which is double the time to the highest point due to the symmetry in projectile motion.
Symmetry in Physics
One beautiful aspect of physics, especially in projectile motion, is symmetry. For a projectile like our football, being fired from and returning to the same height, its path is symmetric. What this means is that the time taken to reach the peak of its trajectory is equal to the time taken to descend back. Hence, the total time of flight is simply twice the time taken to rise to the highest point. In the given problem, the football takes about 1.53 seconds to reach its peak, returning takes the same amount of time, totaling approximately 3.06 seconds.

This symmetry principle allows us to predict behaviors and durations in many projectile scenarios without recalculating the descent phase separately.
Acceleration Due to Gravity
Gravity is a constant force every object with mass experiences, pulling it toward Earth's center. In projectile motion, it plays a crucial role affecting only the vertical component. For the football, as it rises and falls, gravity applies a consistent acceleration of 9.8 m/s² downward.

This means every second, the vertical velocity decreases by 9.8 m/s as it ascends, and increases by the same rate as it descends. Therefore, the total change in vertical velocity from launch to the highest point is 15 m/s, corresponding to the initial vertical speed. Understanding gravity's role in vertical motion helps us calculate key properties of a trajectory, such as the maximum height and time in motion.

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Most popular questions from this chapter

A physics book slides off a horizontal tabletop with a speed of \(1.10 \mathrm{~m} / \mathrm{s} .\) It strikes the floor in \(0.350 \mathrm{~s}\). Ignore air resistance. Find (a) the height of the tabletop above the floor, (b) the horizontal distance from the edge of the table to the point where the book strikes the floor, and (c) the horizontal and vertical components of the book's velocity, and the magnitude and direction of its velocity, just before the book reaches the floor.

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