Problem 105 - Fundamentals of Physics

Question

A projectile is launched with an initial speed of \(30 \mathrm{~m} / \mathrm{s}\) at an angle of \(60^{\circ}\) above the horizontal. What are the (a) magnitude and (b) angle of its velocity \(2.0 \mathrm{~s}\) after launch, and \((\mathrm{c})\) is the angle above or below the horizontal? What are the (d) magnitude and (e) angle of its velocity \(5.0 \mathrm{~s}\) after launch, and \((\mathrm{f})\) is the angle above or below the horizontal?

Step-by-Step Solution

Answer

At 2.0 s, velocity: 16.31 m/s at 23.24° above horizontal. At 5.0 s, velocity: 27.47 m/s at 57.35° below horizontal.

Step 1: Break Down Initial Velocity

Given that the initial speed of the projectile is 30 m/s and the launch angle is 60 degrees, we need to decompose this velocity into horizontal and vertical components.- The horizontal component of the velocity is given by \(v_{0x} = v_0 \cdot \cos(60^{\circ})\).- The vertical component of the velocity is given by \(v_{0y} = v_0 \cdot \sin(60^{\circ})\).Substituting the given values:\[v_{0x} = 30 \cdot \cos(60^{\circ}) = 30 \cdot \frac{1}{2} = 15 \, \text{m/s}\,\]\[v_{0y} = 30 \cdot \sin(60^{\circ}) = 30 \cdot \frac{\sqrt{3}}{2} = 15\sqrt{3} \, \text{m/s}\,\]

Step 2: Determine Velocity Components at 2.0 s

To find the velocity of the projectile 2.0 s after launch, we calculate its horizontal and vertical components at this time.- The horizontal component \(v_x\) remains constant as \(v_{0x} = 15 \, \text{m/s}\).- The vertical component changes due to acceleration due to gravity. It's given by \(v_y = v_{0y} - g \cdot t\). Using \(g = 9.8 \, \text{m/s}^2\):\[v_y = 15\sqrt{3} - 9.8 \cdot 2 = 25.98 - 19.6 = 6.38 \, \text{m/s}\,\]

Step 3: Find Magnitude and Angle at 2.0 s

Using the components from Step 2, calculate the magnitude \(v\) and direction \(\theta\) of the velocity:- The magnitude is given by \(v = \sqrt{v_x^2 + v_y^2}\):\[v = \sqrt{15^2 + 6.38^2} = \sqrt{225 + 40.7044} = \sqrt{265.7044} \approx 16.31 \, \text{m/s}\,\]- The angle with the horizontal is \(\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)\):\[\theta = \tan^{-1}\left(\frac{6.38}{15}\right) \approx 23.24^{\circ}\, \text{above the horizontal}.\]

Step 4: Determine Velocity Components at 5.0 s

Now, calculate the velocity components at 5.0 s:- The horizontal component \(v_x\) is still 15 m/s.- The vertical component \(v_y\) is now:\[v_y = 15\sqrt{3} - 9.8 \cdot 5 = 25.98 - 49 = -23.02 \, \text{m/s}\,\] indicating it is moving downwards.

Step 5: Find Magnitude and Angle at 5.0 s

Calculate the magnitude and direction of the velocity at 5.0 s using the results from Step 4:- The magnitude is:\[v = \sqrt{15^2 + (-23.02)^2} = \sqrt{225 + 529.8804} = \sqrt{754.8804} \approx 27.47 \, \text{m/s}\,\]- The angle with the horizontal is:\[\theta = \tan^{-1}\left(\frac{23.02}{15}\right) \approx 57.35^{\circ}\, \text{below the horizontal}.\]

Step 6: Answer Questions

Based on the calculations: - (a) Magnitude of velocity at 2.0 s: approx. 16.31 m/s - (b) Angle at 2.0 s: approx. 23.24° above the horizontal - (c) The angle is above the horizontal. - (d) Magnitude of velocity at 5.0 s: approx. 27.47 m/s - (e) Angle at 5.0 s: approx. 57.35° - (f) The angle is below the horizontal.

Key Concepts

Initial Velocity Components

When a projectile is launched, its initial velocity can be divided into two components: horizontal and vertical. This division is crucial for understanding the behavior of the projectile throughout its motion. By breaking down the initial velocity, we essentially translate a complex diagonal motion into more manageable horizontal and vertical motions.

• Horizontal Component: This is found using the cosine of the launch angle. For a projectile launched at an angle \( \theta \), the horizontal component of the velocity \( v_{0x} \) is calculated as \( v_{0x} = v_0 \cdot \cos(\theta) \). In our example, with an initial speed \( v_0 = 30 \; \text{m/s} \) and a launch angle of \( 60^{\circ} \), the formula becomes \( v_{0x} = 30 \cdot \cos(60^{\circ}) = 15 \; \text{m/s} \).
• Vertical Component: This is calculated using the sine of the launch angle. It is given by \( v_{0y} = v_0 \cdot \sin(\theta) \). For the given problem, \( v_{0y} = 30 \times \sin(60^{\circ}) = 15\sqrt{3} \; \text{m/s} \).

Understanding these components is essential because they remain independent throughout the projectile's flight, with gravity only affecting the vertical component.

Velocity Magnitude and Direction

Determining the magnitude and direction of a projectile's velocity at any point in time is key to analyzing its motion. After launch, the velocity isn’t just the initial speed; it changes as the projectile moves.

Calculating Magnitude

The magnitude of velocity is derived from the combined effect of horizontal and vertical velocity components. It is computed using the Pythagorean theorem: \( v = \sqrt{v_x^2 + v_y^2} \). For instance, after 2.0 seconds, with \( v_x = 15 \; \text{m/s} \) and \( v_y = 6.38 \; \text{m/s} \), the magnitude is \( \sqrt{15^2 + 6.38^2} \approx 16.31 \; \text{m/s} \).

Determining Direction

The direction at a given time \( \theta \) is the angle made with the horizontal line. It is found using the inverse tangent function: \( \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) \). With our components, after 2.0 seconds, the angle is \( \tan^{-1}\left(\frac{6.38}{15}\right) \approx 23.24^{\circ} \), which is above the horizontal.
Tracking these changes in magnitude and direction helps in predicting the projectile's path and where it will land.

Effect of Gravity on Projectile

Gravity plays a pivotal role in projectile motion by affecting the vertical component of the velocity. Unlike the horizontal component, which remains constant in the absence of air resistance, gravity continuously acts downwards on the projectile.

• Constant Acceleration: Gravity imparts a constant acceleration of \( 9.8 \; \text{m/s}^2 \) downwards. This means the vertical velocity component decreases on the way up, becomes zero at the peak, and then increases in the downward direction.
• Vertical Velocity Change: The formula used to determine the vertical component of the velocity at any time \( t \) is \( v_y = v_{0y} - g \cdot t \). For example, after 5 seconds, given \( v_{0y} = 15\sqrt{3} \; \text{m/s} \), the vertical velocity is \( 15\sqrt{3} - 9.8 \cdot 5 = -23.02 \; \text{m/s} \), indicating it's moving downwards.

Recognizing how gravity affects the projectile helps in understanding its trajectory, especially how high it will go and how long it will stay in the air.

Horizontal and Vertical Components

The separation of motion into horizontal and vertical components simplifies the analysis of projectile motion. These components behave independently, which allows us to solve problems more efficiently.

• Horizontal Motion: Because there is no acceleration acting horizontally (ignoring air resistance), the horizontal component of velocity \( v_x \) remains constant throughout the flight. In this example, \( v_x = 15 \; \text{m/s} \) continues unchanged at both 2.0 seconds and 5.0 seconds.
• Vertical Motion: Affected by gravity, this component changes. At 2.0 seconds, the vertical velocity effects are computed as previously shown. By 5.0 seconds, the downward influence of gravity results in a negative vertical velocity \( v_y = -23.02 \; \text{m/s} \).

By understanding these components, students can predict different stages of the projectile's motion, and how both horizontal and vertical influences guide its journey through the space.