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Chapter 3: Problem 33

(II) A football is kicked at ground level with a spced of 18.0 \(\mathrm{m} / \mathrm{s}\) at an angle of \(38.0^{\circ}\) to the horizontal. How much later does it hit the ground?

Short Answer

Expert verified
The football hits the ground after approximately 2.26 seconds.

Step by step solution

01

Identify the Known Values

We are given the initial speed of the football, \( v_0 = 18.0 \, \mathrm{m/s} \), and the angle \( \theta = 38.0^{\circ} \) from the horizontal.The acceleration due to gravity \( g \) is \( 9.81 \, \mathrm{m/s}^2 \).
02

Resolve Initial Velocity

Resolve the initial velocity into its horizontal and vertical components using trigonometry.\[ v_{0x} = v_0 \cdot \cos(\theta) \] \[ v_{0y} = v_0 \cdot \sin(\theta) \] Calculate these components: \( v_{0x} = 18.0 \cdot \cos(38.0^{\circ}) \) and \( v_{0y} = 18.0 \cdot \sin(38.0^{\circ}) \).
03

Calculate Time to Reach Maximum Height

To find the time to reach maximum height, set the final vertical velocity to zero and use the equation: \[ 0 = v_{0y} - g \cdot t_{up} \] Solving for \( t_{up} \), we have: \[ t_{up} = \frac{v_{0y}}{g} \].
04

Calculate Total Time in Air

The time to ascend to the maximum height is the same as the time to descend back to the original height, so the total time is \( t_{total} = 2 \cdot t_{up} \).
05

Substitute and Solve

Substitute the value of \( v_{0y} \) from Step 2 and \( g \) into the equation for \( t_{up} \): \[ t_{total} = 2 \cdot \frac{18.0 \cdot \sin(38.0^{\circ})}{9.81} \] Calculate \( t_{total} \).
06

Verify the Calculation

Recheck calculations, ensuring that trigonometric functions are correctly evaluated in radians if necessary, and verify that the arithmetic is correct.

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Key Concepts

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

Initial Velocity
In projectile motion, the initial velocity is crucial because it determines how fast and at what direction an object is launched. The initial velocity is denoted by \( v_0 \) and expressed in meters per second (\( \text{m/s} \)). It combines two components: horizontal and vertical velocities that depend on the angle of projection.
  • The horizontal component \( v_{0x} \) can be calculated using \( v_0 \cos(\theta) \).
  • The vertical component \( v_{0y} \) is found using \( v_0 \sin(\theta) \).
These components are necessary to understand how the object will move over time, as they influence both the range and the time of flight.
Angle of Projection
The angle at which a projectile is launched plays a vital role in determining its path. This is called the angle of projection, represented by \( \theta \), and is measured from the horizontal. A key result is that at a \( 45^\circ \) angle, the range of the projectile is maximized under the same initial speed. However, our problem deals with an angle of \( 38^\circ \).
The angle affects:
  • The distribution between vertical and horizontal components of the initial velocity.
  • The height the projectile will reach.
  • The distance it will cover.
In trigonometric terms, this means the sine and cosine of \( \theta \) are used to separate the initial velocity into its useful parts.
Acceleration Due to Gravity
Gravity is a consistent force acting on projectiles, pulling them towards the Earth's surface at a constant acceleration of \( 9.81 \text{ m/s}^2 \). This downward force affects the vertical motion of the projectile.
Here's how gravity comes into play:
  • It causes the vertical component of velocity to decrease as the projectile rises until it momentarily reaches zero at the peak.
  • Then, it accelerates the projectile downward, reinstating its vertical speed as it falls.
Without gravity, objects would travel in a straight line at a constant speed. Therefore, understanding how gravity affects motion is fundamental in predicting projectile behavior.
Time of Flight
The time of flight is the total time the projectile remains in motion from the moment it is launched until it returns to the ground level. This can be divided into two phases: rising to the peak and falling back down.
We find the time to peak using:\[ t_{\text{up}} = \frac{v_{0y}}{g} \]Since the time to rise and the time to fall are symmetrical, the total time of flight is:\[ t_{\text{total}} = 2 \cdot t_{\text{up}} \]This insight helps to determine how long an object stays airborne, allowing for predictions about where it will land. Mastery of the time of flight concept ensures a thorough understanding of the temporal dimension of projectile motion.

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Most popular questions from this chapter

\((a)\) A skier is accelerating down a \(30.0^{\circ}\) hill at \(1.80 \mathrm{~m} / \mathrm{s}^{2}\) (Fig. \(3-39\) ). What is the vertical component of her acceleration? (b) How long will it take her to reach the bottom of the hill, assuming she starts from rest and accelerates uniformly, if the elevation change is \(325 \mathrm{~m} ?\)

A person in the passenger basket of a hot-air balloon throws a ball horizontally outward from the basket with speed \(10.0 \mathrm{~m} / \mathrm{s}\) (Fig. \(3-52\) ). What initial velocity (magnitude and direction) does the ball have relative to a person standing on the ground \((a)\) if the hot-air balloon is rising at \(5.0 \mathrm{~m} / \mathrm{s}\) relative to the ground during this throw, (b) if the hot-air balloon is descending at \(5.0 \mathrm{~m} / \mathrm{s}\) relative to the ground.

The speed of a boat in still water is \(v .\) The boat is to make a round trip in a river whose current travels at speed \(u\). Derive a formula for the time needed to make a round trip of total distance \(D\) if the boat makes the round trip by moving (a) upstream and back downstream, and ( \(b\) ) directly across the river and back. We must assume \(u

If \(V_{x}=7.80\) units and \(V_{y}=-6.40\) units, determine the magnitude and direction of \(\overrightarrow{\mathbf{V}}\).

(II) An airplane is heading due south at a speed of 580 \(\mathrm{km} / \mathrm{h}\) . If a wind begins blowing from the southwest at a speed of 90.0 \(\mathrm{km} / \mathrm{h}\) (average), calculate \((a)\) the velocity (magnitude and dircction) of the plane, relative to the ground, and (b) how far from its intended position it will be after 11.0 \(\mathrm{min}\) if the pilot takes no corrective action.
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