Problem 28 - College Physics

Question

\(\bullet\) Two archers shoot arrows in the same direction from the same place with the same initial speeds but at different angles. One shoots at \(45^{\circ}\) above the horizontal, while the other shoots at \(60.0^{\circ} .\) If the arrow launched at \(45^{\circ}\) lands 225 \(\mathrm{m}\) from the archer, how far apart are the two arrows when they land? (You can assume that the arrows start at essentially ground level.)

Step-by-Step Solution

Answer

The arrows land 25.3 meters apart.

Step 1: Calculate the Initial Velocity

Use the range formula for projectile motion to find the initial velocity \( v_0 \) of the arrow shot at \( 45^{\circ} \). The formula for the range \( R \) of a projectile launched at angle \( \theta \) is \( R = \frac{v_0^2 \sin(2\theta)}{g} \), where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity. Given \( R = 225 \, \text{m} \) and \( \theta = 45^{\circ} \), solve for \( v_0 \).

Step 2: Apply the Range Formula to Solve for Initial Velocity

Substituting the known values into the range formula for the first arrow: \[225 = \frac{v_0^2 \sin(90^{\circ})}{9.8}\]\( \sin(90^{\circ}) = 1 \), so the equation simplifies to:\[v_0^2 = 225 \times 9.8\]Calculating this gives \( v_0^2 = 2205 \, \text{m}^2/\text{s}^2 \). Taking the square root, \( v_0 = 47 \, \text{m/s} \).

Step 3: Calculate the Range for the Second Arrow

Use the same range formula to find the range \( R_2 \) of the arrow shot at \( 60^{\circ} \) with the same initial velocity \( 47 \, \text{m/s} \):\[R_2 = \frac{v_0^2 \sin(2\times60^{\circ})}{g} = \frac{2205 \times \sin(120^{\circ})}{9.8}\]Since \( \sin(120^{\circ}) = \sin(60^{\circ}) = \frac{\sqrt{3}}{2} \), we substitute to get:\[R_2 = \frac{2205 \times \frac{\sqrt{3}}{2}}{9.8}\]Simplifying further gives \( R_2 \approx 199.7 \, \text{m} \).

Step 4: Determine the Distance Between the Arrows When They Land

Find the difference in the range between the two arrows to determine how far apart they are when they land:\[\text{Distance apart} = R_1 - R_2 = 225 \, \text{m} - 199.7 \, \text{m} = 25.3 \, \text{m}\]

Step 5: Conclusion

The two arrows land 25.3 meters apart.

Key Concepts

Range Formula

Projectile motion is a fascinating topic where a projectile is under the influence of gravity and initially propelled at a certain angle and velocity. Understanding the range formula is key when analyzing this type of motion.
The range formula is given by the equation \[ R = \frac{v_0^2 \sin(2\theta)}{g} \]
where:

• \( R \) is the range of the projectile.
• \( v_0 \) is the initial velocity.
• \( \theta \) is the angle of projection.
• \( g \) is the acceleration due to gravity, which is approximately \( 9.8 \, \text{m/s}^2 \).

The formula shows that the range depends on both the square of initial velocity and the sine of twice the angle of projection. This perfect balance can help us calculate how far a projectile travels horizontally when shot from an initial angle above the horizontal. For the case where an arrow is shot at a \(45^\circ\) angle, this maximizes the range since \(\sin(2 \times 45^\circ) = 1\), the highest value the sine function can reach.
When solving problems in projectile motion, using the range formula allows you to confidently determine how far the projectile will land from its starting point, given the right conditions.

Initial Velocity

In the context of projectile motion, the initial velocity \( v_0 \) is an essential component, representing the speed at which a projectile is launched. This is a crucial factor in determining how far or how high an object will go. The initial velocity can have both horizontal and vertical components if the projectile is launched at an angle.
Let's say you are tasked with finding the initial velocity when a projectile is launched. Using the range formula, you can solve for \( v_0 \) if you know the range and angle of projection.
For example, when an arrow is fired with an angle of \(45^\circ\), and lands \(225 \, \text{m}\) away, applying the range formula:\[ 225 = \frac{v_0^2 \times \sin(90^\circ)}{9.8} \]
yields \(v_0 = 47 \, \text{m/s}\) after calculation. This value represents the speed needed at the moment of launch to achieve the given range at specified conditions.
Understanding initial velocity gives you a complete picture of the projection dynamics and aids in predicting how projectiles behave under different conditions.

Angles of Projection

The angle of projection \( \theta \) is a significant factor that affects the trajectory of a projectile in motion. It determines the initial direction and greatly influences how far a projectile will travel. In projectile motion scenarios, even minor adjustments to the angle can have substantial effects on the outcomes.
When projectiles are launched at a \(45^\circ\) angle, they achieve the maximum possible range because this angle optimizes the balance between horizontal and vertical distance traveled. However, different angles can result in varying ranges and heights.
In our scenario, the archers shoot arrows at \(45^\circ\) and \(60^\circ\). For the \(60^\circ\) projection, the arrow is launched more vertically, reducing horizontal range compared to \(45^\circ\), resulting in a shorter range of approximately \(199.7 \, \text{m}\).
When solving projectile motion problems, selecting the best angle requires balancing the need for maximum height versus range, considering that each angle affects the relative trajectories.