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Chapter 3: Problem 33

A football is kicked at ground level with a speed 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 approximately 2.27 seconds later.

Step by step solution

01

Break the Speed into Components

Break the initial speed into its horizontal and vertical components. Use trigonometric functions:- Horizontal component, \( v_{x} = v \cdot \cos(\theta) \) - Vertical component, \( v_{y} = v \cdot \sin(\theta) \) For given values, \( v = 18.0 \) m/s and \( \theta = 38.0^{\circ} \), we find:\[ v_{x} = 18.0 \times \cos(38.0^{\circ}) \approx 14.2 \text{ m/s} \]\[ v_{y} = 18.0 \times \sin(38.0^{\circ}) \approx 11.1 \text{ m/s} \]
02

Calculate the Time of Flight

Use the kinematic equation for vertical motion to calculate the time the ball stays in the air. The equation is:\[ y = v_{y}t + \frac{1}{2} a t^2 \]Since the ball starts and ends at ground level, \( y = 0 \), and acceleration \( a = -g = -9.8 \) m/s虏:\[ 0 = 11.1t - \frac{1}{2} \times 9.8 \times t^2 \]This simplifies to:\[ 0 = 11.1t - 4.9t^2 \]Factor out \( t \):\[ t(11.1 - 4.9t) = 0 \]This gives two solutions; \( t = 0 \) (launch time) and \( t = \frac{11.1}{4.9} \approx 2.27 \text{ s} \). The time of flight is \( 2.27 \text{ s} \).

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Key Concepts

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

Kinematics
Kinematics is a branch of physics that deals with the motion of objects without considering the factors which cause this motion. It focuses on aspects such as displacement, speed, velocity, and acceleration. In our exercise, we deal with projectile motion, which is a common topic in kinematics. Projectile motion refers to the motion of an object thrown or projected into the air, subject to only the acceleration of gravity.

When a football is kicked, like in the original problem, it undergoes projectile motion. This motion is two-dimensional because it involves both horizontal and vertical movement. In kinematics, we break this motion into two components:
- **Horizontal motion**: This component has a constant velocity (if air resistance is ignored) because no acceleration acts horizontally.
- **Vertical motion**: This component has an accelerated velocity because gravity acts downwards, pulling the object towards the earth.

The original problem requires calculating how long the football remains in the air, which involves understanding both the horizontal and vertical components of kinematics.
Trigonometry in Physics
Trigonometry is a branch of mathematics dealing with the relationships between the angles and sides of triangles. In physics, especially in problems involving projectile motion, trigonometry helps us break down complex movements into simpler ones.

In the exercise, the initial speed of the football and the angle at which it is kicked give us two vital components:
  • **Horizontal Component**: Calculated using the cosine function, given as \(v_{x} = v \cdot \cos(\theta)\). This represents how fast the ball travels along the ground.
  • **Vertical Component**: Calculated using the sine function, represented as \(v_{y} = v \cdot \sin(\theta)\). This indicates how quickly the ball rises and falls due to gravity.
Understanding these trigonometric functions is crucial for analyzing the projectile's flight path. By decomposing the initial velocity into horizontal and vertical components, the seemingly complex motion becomes easier to handle and compute.
Time of Flight
Time of flight refers to the total time an object spends in the air from the moment it is launched until it touches the ground again. It is a vital parameter in analyzing projectile motion.

To calculate the time of flight, we focus on vertical motion because the object starts and ends at the same vertical position in these types of problems (ignoring air resistance). Here鈥檚 how it鈥檚 done in the given exercise:
  • Start with the kinematic equation for vertical motion: \(y = v_{y}t + \frac{1}{2} a t^2\). Since the ball hits the ground again at the same height from which it was kicked, \(y = 0\).
  • We know that acceleration \(a = -9.8\, m/s^2\) (due to gravity), so substituting the known values into the equation leads to solving for \(t\).
  • You end up with the equation \(0 = 11.1t - 4.9t^2\), which is solved by factoring out \(t\) leading to values that indicate both the time of launch (\(t = 0\)) and the time when the object lands (\(t \approx 2.27\, s\)).
This solution helps us understand how long the football stays airborne, which is crucial for planning any sports activities involving projectiles.

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Most popular questions from this chapter

(II) An athlete executing a long jump leaves the ground at a \(27.0^{\circ}\) angle and lands 7.80 \(\mathrm{m}\) away. (a) What was the takeoff spced? (b) If this speed were increased by just 5.0\(\%\) , how much longer would the jump be?

A person going for a morning jog on the deck of a cruise ship is running toward the bow (front) of the ship at \(2.0 \mathrm{~m} / \mathrm{s}\) while the ship is moving ahead at \(8.5 \mathrm{~m} / \mathrm{s}\). What is the velocity of the jogger relative to the water? Later, the jogger is moving toward the stern (rear) of the ship. What is the jogger's velocity relative to the water now?

Huck Finn walks at a speed of \(0.70 \mathrm{~m} / \mathrm{s}\) across his raft (that is, he walks perpendicular to the raft's motion relative to the shore). The raft is traveling down the Mississippi River at a speed of \(1.50 \mathrm{~m} / \mathrm{s}\) relative to the river bank (Fig. \(3-49\) ). What is Huck's velocity (speed and direction) relative to the river bank?

Graphically determine the resultant of the following three vector displacements: (1) \(24 \mathrm{~m}, 36^{\circ}\) north of east; (2) \(18 \mathrm{~m}\) \(37^{\circ}\) east of north; and (3) \(26 \mathrm{~m}, 33^{\circ}\) west of south.

You are driving south on a highway at \(25 \mathrm{~m} / \mathrm{s}\) (approximately \(55 \mathrm{mi} / \mathrm{h}\) ) in a snowstorm. When you last stopped, you noticed that the snow was coming down vertically, but it is passing the windows of the moving car at an angle of \(37^{\circ}\) to the horizontal. Estimate the speed of the snowflakes relative to the car and relative to the ground.
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