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\(\bullet\) 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.)

Short Answer

Expert verified
The arrows land 25.3 meters apart.

Step by step solution

01

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 \( v_0 \).
02

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, \( v_0 = 47 \, \text{m/s} \).
03

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 \( R_2 \approx 199.7 \, \text{m} \).
04

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:\[\text{Distance apart} = R_1 - R_2 = 225 \, \text{m} - 199.7 \, \text{m} = 25.3 \, \text{m}\]
05

Conclusion

The two arrows land 25.3 meters apart.

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Key Concepts

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

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.

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Most popular questions from this chapter

\cdotse: Leaping the river, II. A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle. (See Figure \(3.45 .\) ) The takeoff ramp was inclined at \(53.0^{\circ},\) the river was 40.0 \(\mathrm{m}\) wide, and the far bank was 15.0 \(\mathrm{m}\) lower than the top of the ramp. The river itself was 100 \(\mathrm{m}\) below the ramp. You can ignore air resistance. (a) What should his speed have been at the top of the ramp for him to have just made it to the edge of the far bank? (b) If his speed was only half the value found in \((a)\) where did he land?

Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 25.0 \(\mathrm{m} / \mathrm{s}\) as it leaves the hose nozzle. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation of the hose until the water takes 3.00 s to reach a building 45.0 m away. You can ignore air resistance; assume that the end of the hose is at ground level.(a) Find the angle of elevation of the hose. (b) Find the speed and acceleration of the water at the highest point in its trajectory. (c) How high above the ground does the water strike the building, and how fast visit moving just before it hits the building?

A baseball pitcher throws a fastball horizontally at a speed of 42.0 \(\mathrm{m} / \mathrm{s} .\) Ignoring air resistance, how far does the ball drop between the pitcher's mound and home plate, 60 \(\mathrm{ft} 6\) in away?

A major leaguer hits a baseball so that it leaves the bat at a speed of 30.0 \(\mathrm{m} / \mathrm{s}\) and at an angle of \(36.9^{\circ}\) above the horizontal. You can ignore air resistance. (a) At what two times is the base- ball at a height of 10.0 \(\mathrm{m}\) above the point at which it left the bat?(b) Calculate the horizontal and vertical components of the baseball's velocity at each of the two times you found in part (a). (c) What are the magnitude and direction of the base- ball's velocity when it returns to the level at which it left the bat?

A military helicopter on a training mission is flying horizontally at a speed of 60.0\(\mathrm{m} / \mathrm{s}\) when it accidentally drops a bomb(fortunately, not armed) at an elevation of 300 \(\mathrm{m.}\) You can ignore air resistance. (a) How much time is required for the bomb to reach the earth? (b) How far does it travel horizontally while falling? (c) Find the horizontal and vertical components of the bomb's velocity just before it strikes the earth. (d) Draw graphs of the horizontal distance vs. time and the vertical distance vs. time for the bomb's motion. (e) If the velocity of the helicopter remains constant, where is the helicopter when the bomb hits the ground?

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