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A \(60 \mathrm{~kg}\) skier starts from rest at height \(H=20 \mathrm{~m}\) above the end of a ski-jump ramp (Fig. \(8-37\) ) 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 max height of the jump is approximately 4.42 m. (b) The height remains the same if the skier's weight increases.

Step by step solution

01

Analyze the Energy Conversion

The skier begins with gravitational potential energy at height \( H = 20 \text{ m} \). This energy is converted into kinetic energy as the skier moves down the ramp.The initial potential energy is given by:\[PE_{i} = mgh = 60 \text{ kg} \times 9.8 \text{ m/s}^2 \times 20 \text{ m}\]The final kinetic energy at the end of the ramp before the jump:\[KE = \frac{1}{2}mv^2\]Assuming no energy losses, \(PE_{i} = KE\).
02

Solve for Velocity at Ramp End

With \(PE_{i} = KE\), the equation becomes:\[mgh = \frac{1}{2}mv^2\]Solving for velocity \(v\):\[\v^2 = 2gh = 2 \times 9.8 \text{ m/s}^2 \times 20 \text{ m}\]\[\v = \sqrt{2 \times 9.8 \times 20} \approx 19.8 \text{ m/s}\]
03

Determine Vertical Velocity Component

The skier's velocity on leaving the ramp has both horizontal \(v_x\) and vertical \(v_y\) components. Use the initial velocity \(v\) and angle \(\theta = 28^\circ\):\[v_y = v \sin \theta = 19.8 \text{ m/s} \times \sin(28^\circ)\]Calculate \(v_y\):\[v_y = 19.8 \times 0.4695 \approx 9.3 \text{ m/s}\]
04

Calculate Maximum Height h

At the maximum height of the jump, the vertical velocity \(v_y\) becomes zero. Use the conservation of energy or kinematic equation:\[0 = v_y^2 - 2gh \rightarrow h = \frac{v_y^2}{2g}\]Substitute the knowns:\[h = \frac{(9.3 \text{ m/s})^2}{2 \times 9.8 \text{ m/s}^2} \approx 4.42 \text{ m}\]
05

Consider Effects of Increased Weight

Consider the second part of the question: If the skier increases his weight by using a backpack, would the maximum height change?Since the potential energy is converted into kinetic energy, and independent of mass (it cancels out in the energy equations), the maximum height \(h\) reached is the same regardless of the skier's mass.

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Key Concepts

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

kinematic equations
Kinematic equations are essential in the study of motion, allowing us to predict various parameters like the position, velocity, or acceleration of an object over time. When working with the problem of a skier jumping off a ramp, these equations help determine crucial aspects of the skier's flight. To apply kinematic equations effectively, it's vital to differentiate between horizontal and vertical motion.
  • For vertical motion, the equation \(v_y^2 = v_{0y}^2 - 2gh\) is used to find the maximum height \(h\) of the skier above the ramp. \(v_y\) represents the vertical velocity component, and \(v_{0y}\) is the initial vertical velocity.
  • Horizontal motion, not influenced by gravity, simply transfers the horizontal component of velocity unchanged when no air resistance is involved.
To solve for unknowns, identify initial conditions and apply these equations step by step. Remember, constant factors such as acceleration due to gravity \( g \) (\(9.8 \text{ m/s}^2\)) play a massive role in your calculations.
gravitational potential energy
Gravitational potential energy (GPE) is the energy an object possesses due to its position relative to a lower height. It's a foundational concept in physics and incredibly pertinent when tackling problems involving heights, like skiing off ramps. The formula for gravitational potential energy is \(PE = mgh\), where \(m\) is the mass, \(g\) is the gravitational acceleration (\(9.8 \text{ m/s}^2\) on Earth), and \(h\) is the height.
  • When our skier starts at a height \(H = 20 \text{ m}\), all of his energy is gravitational potential energy.
  • As he travels down the frictionless ramp, GPE transforms into kinetic energy (KE), which indicates movement. Therefore, any increase in initial height directly results in an increase in available energy and velocity at the ramp's end.
This energy transformation highlights the conservation of energy principle. No energy is lost in a frictionless system, allowing us to equate initial potential energy to final kinetic energy to solve for parameters like velocity at the bottom of the ramp.
projectile motion
Projectile motion is a type of two-dimensional motion that involves an object being launched into the air and affected only by gravity (ignoring air resistance). In our skier's scenario, the projectile motion takes place as he leaves the ramp at an angle \(\theta = 28^{\circ}\). Understanding this motion involves:
  • Breaking down the skier's velocity into horizontal and vertical components. The horizontal component \(v_x = v \cos \theta\) and the vertical component \(v_y = v \sin \theta\).
  • Using these components to predict the path and maximum height of the projectile, a task simplified by recognizing that the vertical component dictates height, while the horizontal influences range.
By calculating how high the skier goes, we determine critical points of his trajectory, such as the maximum height. The constant effect of gravity eventually brings him back down, unaffected by his speed or the angle of launch when they are combined in the energy equation. Thus, understanding projectile motion offers insight into the predictability and nature of two-dimensional motion governed by gravity.

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Most popular questions from this chapter

A block with mass \(m=2.00 \mathrm{~kg}\) is placed against a spring on a frictionless incline with angle \(\theta=30.0^{\circ}\) (Fig. 8-44). (The block is not attached to the spring.) The spring, with spring constant \(k=19.6 \mathrm{~N} / \mathrm{cm},\) is compressed \(20.0 \mathrm{~cm}\) and then released. (a) What is the elastic potential energy of the compressed spring? (b) What is the change in the gravitational potential energy of the block- Earth system as the block moves from the release point to its highest point on the incline? (c) How far along the incline is the highest point from the release point?

A \(2.50 \mathrm{~kg}\) beverage can is thrown directly downward from a height of \(4.00 \mathrm{~m},\) with an initial speed of \(3.00 \mathrm{~m} / \mathrm{s}\). The air drag on the can is negligible. What is the kinetic energy of the can (a) as it reaches the ground at the end of its fall and (b) when it is halfway to the ground? What are (c) the kinetic energy of the can and (d) the gravitational potential energy of the can-Earth system \(0.200 \mathrm{~s}\) before the can reaches the ground? For the latter, take the reference point \(y=0\) to be at the ground.

A factory worker accidentally releases a \(180 \mathrm{~kg}\) crate that was being held at rest at the top of a ramp that is \(3.7 \mathrm{~m}\) long and inclined at \(39^{\circ}\) to the horizontal. The coefficient of kinetic friction between the crate and the ramp, and between the crate and the horizontal factory floor, is \(0.28 .\) (a) How fast is the crate moving as it reaches the bottom of the ramp? (b) How far will it subsequently slide across the floor? (Assume that the crate's kinetic energy does not change as it moves from the ramp onto the floor.) (c) Do the answers to (a) and (b) increase, decrease, or remain the same if we halve the mass of the crate?

A stone with a weight of \(5.29 \mathrm{~N}\) is launched vertically from ground level with an initial speed of \(20.0 \mathrm{~m} / \mathrm{s},\) and the air drag on it is \(0.265 \mathrm{~N}\) throughout the flight. What are (a) the maximum height reached by the stone and (b) its speed just before it hits the ground?

A constant horizontal force moves a \(50 \mathrm{~kg}\) trunk \(6.0 \mathrm{~m}\) up a \(30^{\circ}\) incline at constant speed. The coefficient of kinetic friction is \(0.20 .\) What are (a) the work done by the applied force and (b) the increase in the thermal energy of the trunk and incline?

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