Chapter 4: Problem 105
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?
Short Answer
Step by step solution
Break Down Initial Velocity
Determine Velocity Components at 2.0 s
Find Magnitude and Angle at 2.0 s
Determine Velocity Components at 5.0 s
Find Magnitude and Angle at 5.0 s
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial Velocity Components
- 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, \( v_{0y} = 30 \times \sin(60^{\circ}) = 15\sqrt{3} \; \text{m/s} \).
Velocity Magnitude and Direction
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
- 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.
Horizontal and Vertical Components
- 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 \( v_y = -23.02 \; \text{m/s} \).