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Chapter 8: Problem 22

A \(70 \mathrm{~kg}\) skier starts from rest at height \(H=22 \mathrm{~m}\) above the end of a ski-jump ramp (Fig. 8-25) and leaves the ramp at angle \(\theta=\) \(28^{\circ}\). Neglect the effects of air resistance and assume the ramp is frictionless. (a) What is the maximum height \(h\) of his jump above the end of the ramp? (b) If he increased his weight by putting on a backpack, would \(h\) then be greater, less, or the same?

Short Answer

Expert verified
(a) The maximum height above the end of the ramp is approximately 4.85 m. (b) Adding weight does not change the maximum height.

Step by step solution

01

Calculate the initial potential energy

The skier starts from rest at a height of 22 m. The initial potential energy (PE) is calculated using the formula: \[ PE = mgh \] where \( m = 70 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), and \( h = 22 \text{ m} \). Substitute the values:\[ PE = 70 \times 9.8 \times 22 = 15092 \text{ J} \].
02

Calculate the skier's speed at end of ramp

As the skier reaches the end of the ramp, the entire potential energy converts into kinetic energy (since there is no friction or air resistance). The kinetic energy (KE) at the end of the ramp is:\[ KE = \frac{1}{2}mv^2 \]Equating PE and KE gives:\[ 15092 = \frac{1}{2}\times70\times v^2\]Solve for \(v\):\[ v^2 = \frac{2\times15092}{70} \approx 431.2 \]\[ v \approx \sqrt{431.2} \approx 20.76 \text{ m/s} \].
03

Determine the vertical component of velocity

The skier leaves the ramp at an angle \(\theta = 28^{\circ}\). The vertical component of velocity \(v_y\) can be found using:\[ v_y = v \sin \theta \]Substitute \(v = 20.76 \text{ m/s} \) and \( \theta = 28^{\circ} \):\[ v_y = 20.76 \sin 28^{\circ} \approx 9.75 \text{ m/s} \].
04

Calculate the maximum height of the jump above the ramp

At the maximum height, the vertical component of the skier's velocity is 0. Use the formula for vertical motion:\[ v_y^2 = 2gh \]Rearrange to find \( h \):\[ h = \frac{v_y^2}{2g} \]Substitute \( v_y = 9.75 \text{ m/s} \) and \( g = 9.8 \text{ m/s}^2 \):\[ h = \frac{9.75^2}{2 \times 9.8} \approx 4.85 \text{ m} \].
05

Analyze impact of increased weight on maximum height

The maximum height reached is determined by the vertical component of velocity and gravitational acceleration, neither of which depends on mass in a frictionless scenario. Therefore, increasing the skier's weight by adding a backpack does not affect the maximum height reached.

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Key Concepts

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

Potential Energy
Potential energy is the stored energy of an object due to its position in a gravitational field. In our case, it's the energy stored by the skier at the top of the ski jump ramp. The higher the skier starts from, the more potential energy they have. This potential energy is calculated using the formula:
  • \( PE = mgh \)
where:
  • \( m \) is the mass of the skier, 70 kg in this scenario.
  • \( g \) is the acceleration due to gravity \( (9.8 \, ext{m/s}^2) \).
  • \( h \) is the height from which the skier starts, 22 meters.
By substituting these values into the formula, the potential energy of the skier at the beginning is determined to be 15092 Joules. This value represents the total energy available to be converted into kinetic energy as the skier descends.
Kinetic Energy
Kinetic energy is the energy of motion. As the skier descends the ramp, the potential energy converts to kinetic energy. This energy is calculated by the formula:
  • \( KE = \frac{1}{2} mv^2 \)
where \( v \) is the velocity of the skier at the end of the ramp. Conservation of energy principles dictate that all the potential energy becomes kinetic energy at the ramp's end (in a frictionless context). Therefore,
  • \( 15092 = \frac{1}{2} \times 70 \times v^2 \)
Solving this equation results in a skier velocity of \( 20.76 \, ext{m/s} \) as they leave the ramp. This speed is essential for determining subsequent vertical motion.
Projectile Motion
Projectile motion describes the path of an object launched into the air, moving under the influence of gravity. When the skier leaves the ramp, they follow a parabolic motion. Importantly, this motion can be dissected into horizontal and vertical components. Using trigonometric functions, we calculate these components:
  • Horizontal component: \( v_x = v \cos \theta \)
  • Vertical component: \( v_y = v \sin \theta \)
Given the skier's velocity \( v = 20.76 \, ext{m/s} \) and launch angle \( \theta = 28^{\circ} \):
  • \( v_y = 20.76 \times \sin 28^{\circ} \approx 9.75 \, ext{m/s} \)
The vertical motion component decides the maximum height achieved, whereas the horizontal component determines how far the skier will travel.
Vertical Motion
Vertical motion focuses on the skier's movement against gravity. For the skier's jump, the vertical motion determines the maximum jump height. We know that at the jump's peak, the vertical velocity is 0. The calculation involves the vertical speed \( v_y \) achieving this zero at the pinnacle:
  • \( v_y^2 = 2gh \)
Rearranging to solve for \( h \), the additional height over the ramp that the skier can reach:
  • \( h = \frac{(9.75)^2}{2 \times 9.8} \approx 4.85 \, ext{m} \)
Notably, the mass does not influence the height due to the dependence solely on gravitational acceleration and initial vertical velocity. Thus, carrying a backpack would not alter the outcome in this idealized scenario.

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Most popular questions from this chapter

A \(700 \mathrm{~g}\) block is released from rest at height \(h_{0}\) above a vertical spring with spring constant \(k=450 \mathrm{~N} / \mathrm{m}\) and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring \(19.0 \mathrm{~cm}\). How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of \(h_{0} ?\) (d) If the block were released from height \(2.00 h_{0}\) above the spring, what would be the maximum compression of the spring?

You drop a \(2.00 \mathrm{~kg}\) book to a friend who stands on the ground at distance \(D=10.0 \mathrm{~m}\) below. Your friend's outstretchedhands are at distance \(d=1.50 \mathrm{~m}\) above the ground (Fig. 8-19). (a) How much work \(W_{g}\) does the gravitational force do on the book as it drops to her hands? (b) What is the change \(\Delta U\) in the gravitational potential energy of the book-Earth system during the drop? If the gravitational potential energy \(U\) of that system is taken to be zero at ground level, what is \(U\) (c) when the book is released, and (d) when it reaches her hands? Now take \(U\) to be \(100 \mathrm{~J}\) at ground level and again find (e) \(W_{g}\), (f) \(\Delta U\), (g) \(U\) at the release point, and (h) \(U\) at her hands.

Two children are playing a game in which they try to hit a small box on the floor with a marble fired from a spring-loaded gun that is mounted on a table. The target box is horizontal distance \(D=2.20 \mathrm{~m}\) from the edge of the table; see Fig. 8-36. Bobby compresses the spring \(1.10 \mathrm{~cm}\), but the center of the marble falls \(26.3 \mathrm{~cm}\) short of the center of the box. How far should Rhoda compress the spring to score a direct hit? Assume that neither the spring nor the ball encounters friction in the gun.

A worker pushed a \(23 \mathrm{~kg}\) block \(8.4 \mathrm{~m}\) along a level floor at constant speed with a force directed \(32^{\circ}\) below the horizontal. If the coefficient of kinetic friction between block and floor was \(0.20\), what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block - floor system?

A swimmer moves through the water at an average speed of \(0.22 \mathrm{~m} / \mathrm{s}\). The average drag force is \(110 \mathrm{~N}\). What average power is required of the swimmer?
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