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Chapter 3: Problem 54

(II) An athlete executing a long jump leaves the ground at a \(27.0^{\circ}\) angle and lands 7.80 \(\mathrm{m}\) away. (a) What was the takeoff spced? (b) If this speed were increased by just 5.0\(\%\) , how much longer would the jump be?

Short Answer

Expert verified
The takeoff speed was about 8.53 m/s. With a 5% speed increase, the jump would be about 0.80 m longer.

Step by step solution

01

Understand the Problem

We need to determine the initial takeoff speed of an athlete who jumps at a \(27^{\circ}\) angle and lands 7.80 meters away. Then, we need to calculate how much longer the jump would be if the speed increased by 5\%.
02

Identify the Relevant Formula

For projectile motion, the range \( R \) is given by the formula: \[ R = \frac{v_0^2 \sin(2\theta)}{g} \]where \(v_0\) is the initial velocity, \(\theta\) is the angle of takeoff, and \(g\) is the acceleration due to gravity (approximately \(9.81 \text{ m/s}^2\)).
03

Solve for the Takeoff Speed

We have \( R = 7.80 \) m and \( \theta = 27^{\circ} \). Let's solve for \( v_0 \):\[7.80 = \frac{v_0^2 \sin(54^{\circ})}{9.81}\]Rearrange the formula to solve for \( v_0 \):\[v_0 = \sqrt{\frac{7.80 \times 9.81}{\sin(54^{\circ})}}\]Compute the value:\[v_0 \approx 8.53 \text{ m/s}\]
04

Calculate the Increased Speed

If the initial speed increases by 5\%, the new speed \( v_{0\_new} \) is:\[v_{0\_new} = 8.53 \times 1.05 \approx 8.96 \text{ m/s}\]
05

Determine New Range with Increased Speed

Use the same range formula with the new speed:\[R_{\text{new}} = \frac{8.96^2 \times \sin(54^{\circ})}{9.81}\]Calculate the new range:\[R_{\text{new}} \approx 8.60 \text{ m} \]
06

Calculate Increase in Jump Distance

The increase in the jump distance is:\[\text{Increase} = R_{\text{new}} - 7.80\]\[\text{Increase} \approx 8.60 - 7.80 = 0.80 \text{ m}\]
07

Summary

The original takeoff speed is approximately 8.53 m/s. If the speed is increased by 5%, the jump distance increases by approximately 0.80 meters.

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Key Concepts

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

Initial Velocity
When a projectile is launched, the initial velocity is the speed at which it starts its motion. For the athlete performing a long jump, this is the speed obtained during takeoff. Initial velocity determines how far and how high the projectile will travel.
The initial velocity is essential in calculating other parameters of projectile motion, such as range and height. In the long jump exercise, you're asked to find the initial velocity to calculate how the athlete managed to jump a distance of 7.80 meters.
To find this initial velocity, you use the formula for the range of a projectile:
  • Use the given range (7.80 meters) and the angle of projection (27 degrees).
  • Plug these into the formula for projectile range to solve for the initial velocity.
  • The formula rearranges so that you can solve for velocity: \[v_0 = \sqrt{\frac{R \times g}{\sin(2\theta)}}\]
By solving this equation, you find that the takeoff speed was approximately 8.53 m/s.
Range of Projectile
The range of a projectile is the horizontal distance it travels while in motion. In this case, it's the distance the athlete covers during a long jump.
This distance can be influenced by several key factors:
  • Initial velocity
  • Angle of projection
  • Gravity
Using the range formula, we can understand how these factors interact:\[ R = \frac{v_0^2 \sin(2\theta)}{g} \]In the exercise, the athlete lands 7.80 meters away鈥攖his is the calculated range of the jump. If the initial velocity increases, the range typically increases as well. An increase in velocity by 5% showed that the jump could be 0.80 meters longer. The range is a reflection of how well the initial conditions harness gravity and angle.
Angle of Projection
The angle at which the athlete takes off affects the trajectory and the range of the jump. The angle of projection is crucial in optimizing the conditions for a longer jump. Generally,
  • The angle of 45 degrees is often considered optimal for maximum range without air resistance.
  • In this scenario, the athlete uses an angle of 27 degrees.
  • This angle helps split the initial velocity into vertical and horizontal components.
By understanding the angle's role, one can adjust it alongside velocity to potentially improve performance. It affects not only how far but also how high the jump reaches, impacting factors such as jump trajectory and time in the air.
Acceleration Due to Gravity
Gravity is the force pulling the projectile back to the ground. It plays a fundamental role in projectile motion by affecting how quickly the projectile descends.
The standard acceleration due to gravity on Earth is approximately 9.81 m/s虏. This value is used to find other parameters of projectile motion.
  • Gravity ensures that all projectile motion follows a parabolic path.
  • It affects the duration the projectile stays in the air.
  • The range formula explicitly uses gravity to determine how far a projectile will go based on initial conditions.
Without gravity, projectiles would travel in a straight line forever, without landing. In any calculation of projectile motion, gravity will always be a key component dictating trajectory.

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