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The current world-record motorcycle jump is \(77,0 \mathrm{~m}\), set by Jason Renie. Assume that he left the take-off ramp at \(12.0^{\circ}\) to the horizontal and that the take-off and landing heights are the same. Neglecting air drag, determine his take-off speed.

Short Answer

Expert verified
The take-off speed is approximately 43.1 m/s.

Step by step solution

01

Identify the Known Values

We know that the horizontal distance (range) of the jump is \( R = 77.0 \) meters, and the angle of projection \( \theta = 12.0^{\circ} \). We also know that the landing height is the same as the take-off height, and we need to find the initial speed \( v_0 \).
02

Use the Range Formula

The range \( R \) of a projectile is given by the formula: \[ R = \frac{v_0^2 \sin(2\theta)}{g} \] where \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity.
03

Rearrange for Initial Velocity

Rearrange the range formula to solve for the initial velocity \( v_0 \): \[ v_0 = \sqrt{\frac{R \cdot g}{\sin(2\theta)}} \]
04

Substitute the Values

Substitute the known values into the formula: \( R = 77.0 \) meters, \( g = 9.8 \text{ m/s}^2 \), and \( \theta = 12.0^{\circ} \). Compute \( \sin(2\theta) = \sin(24^{\circ}) \).
05

Calculate the Intermediate Results

Calculate \( \sin(24^{\circ}) \) using a calculator to get approximately 0.4067. Now substitute into the equation: \[ v_0 = \sqrt{\frac{77.0 \cdot 9.8}{0.4067}} \].
06

Compute the Take-off Speed

Perform the calculations in order: Calculate \( 77.0 \times 9.8 = 754.6 \). Divide by \( 0.4067 \) to get approximately \( 1855.5 \). Finally, take the square root to find \( v_0 \approx 43.1 \text{ m/s} \).

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Key Concepts

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

Range Formula
Projectile motion often involves the challenge of calculating the range of a projectile based on given parameters. The range formula specifically helps in determining how far a projectile can travel horizontally, given the initial speed and angle of projection. This is incredibly useful in predicting distances in sports scenarios, like jumping or throwing.

The formula for the range, or horizontal distance (\(R\)), of a projectile, is:
  • \[ R = \frac{v_0^2 \sin(2\theta)}{g} \]
Here,
  • \(v_0\) is the initial velocity (speed).
  • \(\theta\) is the angle of projection.
  • \(g\) is the acceleration due to gravity, usually \(9.8\, \text{m/s}^{2}\) on Earth.
The range formula is derived from the projectile motion equations for horizontal and vertical displacement, and is especially valid when the takeoff and landing heights are the same. The condition \(\sin(2\theta)\) reveals that the range also depends on the angle, most efficient at an angle of \(45^{\circ}\) where the maximal value of \(\sin(2\theta)\) occurs. Students often use this formula specifically for cases where air resistance can be ignored.
Initial Velocity
The initial velocity, denoted as \(v_0\), is crucial in determining the projectile's range. In essence, this is the speed at which the object starts its motion. Calculating the initial velocity requires an understanding of both the range and the effect of projected angle and gravity.

To derive the initial velocity from the range formula, you rearrange it to:
  • \[ v_0 = \sqrt{\frac{R \cdot g}{\sin(2\theta)}} \]
This rearrangement gives us an equation that shows how the range is scaled by gravity and angle of projection to provide the initial velocity. A larger initial velocity suggests that a projectile can cover significant distances, assuming angle and other factors remain constant. For practical situations like jumps or launches, this calculation is key to ensure objects meet their targeted range.
Angle of Projection
The angle of projection (\(\theta\)) is another fundamental aspect of projectile motion. It influences how high and far a projectile travels. In the context of the range formula, the angle affects the scale of the projectile’s displacement both vertically and horizontally.

The effectiveness of the angle arises from the term \(\sin(2\theta)\) in the range formula, which adjusts the initial velocity's impact to maximize or minimize the projectile’s distance. A significant observation is that the range is maximized when \(\theta\) is \(45^{\circ}\). Such an angle provides an optimal balance between horizontal and vertical velocities.

For jumps or throws not aiming for maximum range, angles lower or higher than \(45^{\circ}\) could be used, depending on the desired height or distance. Understanding the role of the angle helps in determining optimal starting angles in real-world scenarios, such as sports, where adjusting this angle can significantly influence performance and safety.

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Most popular questions from this chapter

A rifle that shoots bullets at \(460 \mathrm{~m} / \mathrm{s}\) is to be aimed at a target \(45.7 \mathrm{~m}\) away. If the center of the target is level with the rifle, how high above the target must the rifle barrel be pointed so that the bullet hits dead center?

A dart is thrown horizontally with an initial speed of \(10 \mathrm{~m} / \mathrm{s}\) toward point \(P\), the bull's-eye on a dart board. It hits at point \(Q\) on the rim, vertically below \(P, 0.19 \mathrm{~s}\) later. (a) What is the distance \(P Q ?\) (b) How far away from the dart board is the dart released?

A small ball rolls horizontally off the edge of a tabletop that is \(1.20 \mathrm{~m}\) high. It strikes the floor at a point \(1.52 \mathrm{~m}\) horizontally from the table edge. (a) How long is the ball in the air? (b) What is its speed at the instant it leaves the table?

Long flights at midlatitudes in the Northern Hemisphere encounter the jet stream, an eastward airflow that can affect a plane's speed relative to Earth's surface. If a pilot maintains a certain speed relative to the air (the plane's airspeed), the speed relative to the surface (the plane's ground speed) is more when the flight is in the direction of the jet stream and less when the flight is opposite the jet stream. Suppose a round-trip flight is scheduled between two cities separated by \(4000 \mathrm{~km}\), with the outgoing flight in the direction of the jet stream and the return flight opposite it. The airline computer advises an airspeed of \(1000 \mathrm{~km} / \mathrm{h}\), for which the difference in flight times for the outgoing and return flights is \(70.0\) min. What jet-stream speed is the computer using?

A particle moves along a circular path over a horizontal \(x y\) coordinate system, at constant speed. At time \(t_{1}=4.00 \mathrm{~s}\), it is at point \((5.00 \mathrm{~m}, 6.00 \mathrm{~m})\) with velocity \((3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) and acceleration in the positive \(x\) direction. At time \(t_{2}=10.0 \mathrm{~s}\), it has velocity \((-3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}\) and acceleration in the positive \(y\) direction. What are the (a) \(x\) and (b) \(y\) coordinates of the center of the circular path if \(t_{2}-t_{1}\) is less than one period?

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