91Ó°ÊÓ

As drawn in Fig. 2-7, a ball is thrown from the top of one building toward a tall building \(50 \mathrm{~m}\) away. The initial velocity of the ball is \(20 \mathrm{~m} / \mathrm{s}-40^{\circ}\) above horizontal. How far above or below its original level will the ball strike the opposite wall? We have $$ \begin{array}{l} v_{i x}=(20 \mathrm{~m} / \mathrm{s}) \cos 40^{\circ}=15.3 \mathrm{~m} / \mathrm{s} \\ v_{i x}=(20 \mathrm{~m} / \mathrm{s}) \cos 40^{\circ}=12.9 \mathrm{~m} / \mathrm{s} \end{array} $$ Consider first the horizontal motion. For it, $$ v_{i x}=v_{f x}=v_{x} 15.3 \mathrm{~m} / \mathrm{s} $$ Then \(x=v_{x} t\) gives $$ 50 \mathrm{~m}=(15.3 \mathrm{~m} / \mathrm{s}) t \quad \text { or } \quad t=3.27 \mathrm{~s} $$ For the vertical motion, taking down as positive, $$ y=v_{i j} t+\frac{1}{2} a_{y} t^{2}=(-12.9 \mathrm{~m} / \mathrm{s})(3.27 \mathrm{~s})+\frac{1}{2}\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(3.27 \mathrm{~s})^{2}=10.3 \mathrm{~m} $$ and to two significant figures, \(y=10 \mathrm{~m}\). Since \(y\) is positive, and since down is positive, the ball will hit at \(10 \mathrm{~m}\) below the original level.

Step by step solution

01

Identify the Known Horizontal Components

We start by identifying the horizontal component of the initial velocity. Given the initial velocity of the ball is 20 m/s at an angle of 40° above the horizontal, the horizontal component is calculated using: \( v_{i x} = v_i \cos \theta = (20 \, \text{m/s}) \cos 40^\circ = 15.3 \, \text{m/s} \).

02

Identify the Known Vertical Components

Next, we identify the vertical component of the initial velocity. Use the following equation: \( v_{i y} = v_i \sin \theta = (20 \, \text{m/s}) \sin 40^\circ = 12.9 \, \text{m/s} \). Note that the sign is negative because the vertical motion is being considered downwards as positive.

03

Determine Time of Flight

Considering horizontal motion and knowing the ball is thrown across a 50 m wide gap, use the equation: \( x = v_{x} t \) to find time \( t \). Substitute the known values: \( 50 \, \text{m} = (15.3 \, \text{m/s}) t \), solving for \( t \) gives: \( t = 3.27 \, \text{s} \).

04

Calculate Vertical Displacement

Now consider the vertical motion. Use the equation for vertical displacement: \( y = v_{i y} t + \frac{1}{2} a_{y} t^2 \). Substitute \( v_{i y} = -12.9 \, \text{m/s} \), \( a_y = 9.81 \, \text{m/s}^2 \), and \( t = 3.27 \, \text{s} \): \( y = (-12.9 \, \text{m/s})(3.27 \, \text{s}) + \frac{1}{2}(9.81 \, \text{m/s}^2)(3.27 \, \text{s})^2 = 10.3 \, \text{m} \).

05

Interpret the Result

Since \( y = 10.3 \, \text{m} \) is positive and was calculated with downward motion considered positive, the ball strikes the opposite building 10 meters below its original level.

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Key Concepts

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

horizontal motion

In projectile motion, understanding horizontal motion is crucial. When a projectile is launched, its horizontal component is independent of its vertical component.
The key aspect to remember is that the horizontal velocity remains constant throughout the flight.
This is because, in the absence of air resistance, there are no horizontal forces acting on the projectile, meaning that the projectile's horizontal acceleration is zero.
Hence, the horizontal motion can be described by the simple equation:

• \( x = v_{x} t \)

Where \( x \) is the horizontal distance traveled, \( v_x \) is the initial horizontal velocity, and \( t \) is the time.
This equation represents that the displacement in horizontal motion is directly proportional to time, with the constant of proportionality being the horizontal velocity.
For instance, if our projectile is to cross a horizontal distance of 50 m with a constant horizontal velocity of 15.3 m/s, we utilize the equation \( 50 \, \text{m} = (15.3 \, \text{m/s}) t \) to find the time duration, which is 3.27 seconds.

vertical motion

Vertical motion in projectile scenarios is influenced by gravity, making it quite dynamic compared to horizontal motion.
The vertical component is affected by acceleration due to gravity, which we denote as \( g = 9.81 \, \text{m/s}^2 \).
This means that as the projectile moves upwards, it decelerates, and on its way down, it accelerates. In this sense,
we can determine the vertical displacement using the formula:

• \( y = v_{i y} t + \frac{1}{2} a_{y} t^2 \)

In this equation, \( y \) represents the vertical displacement, \( v_{i y} \) is the initial vertical velocity, and \( a_y \) is the acceleration of gravity, which is always acting downward.
In the given exercise, considering the ball's launch, we used this formula to determine how the ball's position changes vertically and found it goes 10.3 meters below its starting height when it hits the wall.

initial velocity

Initial velocity is a critical factor in projectile motion as it determines how the projectile will travel.
It consists of two components: horizontal and vertical, which can be calculated using trigonometric functions if the launch angle is known.
The process involves the following calculations:

• The horizontal component: \( v_{i x} = v_i \cos \theta \)
• The vertical component: \( v_{i y} = v_i \sin \theta \)

Where \( v_i \) is the initial speed and \( \theta \) is the angle above the horizontal.
For example, in the exercise, we have an initial velocity of 20 m/s at an angle of 40°.
This results in a horizontal component of 15.3 m/s and a vertical component of 12.9 m/s.
Such a separation allows us to individually address horizontal and vertical motions, building a complete picture of the projectile's path.

time of flight

Time of flight is the duration for which a projectile remains in motion until it reaches a specific point.
It helps in determining both how long the projectile will be traveling and where it will land.
The time of flight can be found from either the horizontal or vertical motion analysis.
For horizontal motion, since the horizontal velocity is constant, it helps us easily find time by rearranging the equation \( x = v_{x} t \).
For example, in the exercise, knowing that the ball must travel 50 m horizontally with a constant velocity of 15.3 m/s,
we solve the equation to find the time of flight, which is approximately 3.27 seconds.
This time is crucial for further calculations regarding the vertical displacement, helping in predicting key outcomes of the projectile's flight.