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A bottle rocket can shoot its projectile vertically to a height of 25.0 \(\mathrm{m}\) . At what angle should the bottle rocket be fired to reach its maximum horizontal range, and what is that range? (You can ignore air resistance.)

Short Answer

Expert verified
Fire at 45 degrees for a range of 50.1 m.

Step by step solution

01

Understanding the Concept

The angle for maximum horizontal range in projectile motion is achieved when the projectile is launched at an angle of 45 degrees. This is a fundamental result from physics, as the range is maximized when both the components of initial velocity (horizontal and vertical) are equal.
02

Finding Initial Velocity for Maximum Height

To find the initial velocity needed to reach a maximum height of 25.0 m, we'll use the formula for the maximum height of a projectile given by \[ h = \frac{v_i^2 \sin^2(\theta)}{2g} \]Given that the angle \( \theta \) for maximum height is 90 degrees (since we are only considering the vertical component for initial velocity here), we can simplify to \[ 25.0 = \frac{v_i^2}{2 \times 9.8} \]Solving for \( v_i \) gives \[ v_i = \sqrt{2 \times 9.8 \times 25.0} \approx 22.14 \, \text{m/s} \]
03

Calculating Horizontal Range at 45 Degrees

Now that we know the initial velocity, we can calculate the horizontal range when the projectile is launched at an angle of 45 degrees using \[ R = \frac{v_i^2 \sin(2\theta)}{g} \]Substitute \( \theta = 45 \) degrees and \( v_i = 22.14 \, \text{m/s} \):\[ R = \frac{(22.14)^2 \sin(90)}{9.8} \approx \frac{490.89}{9.8} \approx 50.1 \, \text{m} \]
04

Conclusion

Thus, to achieve the maximum horizontal range, the bottle rocket should be fired at an angle of 45 degrees, and the range will be approximately 50.1 meters.

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Key Concepts

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

Maximum Range
In the realm of projectile motion, the concept of maximum range is crucial. The horizontal distance that an object travels is known as the range. When discussing maximum range, we mean the farthest horizontal distance the projectile can reach under ideal conditions—the optimal balance between elevation and trajectory angle. To achieve the maximum range, the launch angle of the projectile must be precisely 45 degrees. This angle evenly splits the initial velocity into horizontal and vertical components, allowing the projectile to harness gravity's pull effectively while covering the longest possible horizontal stretch.
This is derived from the fact that when the launch angle is 45 degrees:
  • The upward force of gravity's reduction in vertical velocity is matched by the momentum gained in horizontal displacement.
  • Both horizontal and vertical velocities play an equal part in maximizing the distance.
  • The launch angle becomes the key factor, complemented by initial velocity and neglecting resistance factors like air drag.
Initial Velocity
Initial velocity, denoted as \( v_i \), is a critical aspect of projectile motion that influences both height and range of a projectile. In our case, to calculate this velocity for the bottle rocket, aimed at reaching a maximum height of 25 meters, we used the formula:\[ h = \frac{v_i^2 \sin^2(\theta)}{2g}\]
With a 90-degree angle focusing solely on vertical velocity, the initial velocity becomes solely a function of peak height. Solving for \( v_i \), we found an initial velocity of about 22.14 m/s necessary for the projectile to climb 25 meters.
This initial velocity:
  • Sets the stage for subsequent horizontal range calculations.
  • Forms the basis for determining trajectory and ultimate distance traveled.
  • Serves as the starting kinetic energy point for projectiles launched at varying angles.
Projectile Angle
The angle at which a projectile is launched significantly impacts its motion and resultant path. In maximizing horizontal range, the optimal launch angle is typically 45 degrees, where the resultant velocities optimize distance. However, variations in angle directly affect other outcomes:
  • A steep angle (approaching 90 degrees) boosts vertical climb at the expense of horizontal spread.
  • A shallow angle (close to 0 degrees) translates into greater horizontal movement but sacrifices height.
  • Thus, identifying and understanding the right angle is fundamental for tasks like maximizing range or reaching specific heights.

The physics behind this defines the balance of mechanical forces acting on the projectile to precision.

Practical Considerations

In real-world scenarios, considerations such as air resistance, launch altitude, and projectile design can alter the optimal angle and should be accounted for to refine predictions and outcomes.
Physics Problem Solving
Solving physics problems involves understanding variables at play and how they interconnect. The scenario with the bottle rocket is a perfect example illustrating projectile motion principles:
  • Start by identifying known values: such as desired height, launch angle, and environmental constants like gravity.
  • Apply the relevant physics formulas: Use height and range equations to analyze projectile paths.
  • Calculate each component methodically: Dissecting each part of the problem (height, initial velocity, angle) ensures greater accuracy.
Mathematical precision and a logical approach are your best allies to ensure correct and efficient problem solving. Additionally, practice by repeating similar exercises can reinforce these concepts, allowing for a deeper understanding of underlying physics principles and enhancing skills for tackling varied projectile motion problems.

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