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Chapter 3: Problem 45

After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced \(51.0 \mathrm{m}\) horizontally from the end of the ramp. His velocity, just before landing, is \(23.0 \mathrm{m} / \mathrm{s}\) and points in a direction \(43.0^{\circ}\) below the horizontal. Neglecting air resistance and any lift he experiences while airborne, find his initial velocity (magnitude and direction) when he left the end of the ramp. Express the direction as an angle relative to the horizontal.

Short Answer

Expert verified
Initial velocity: magnitude 17.5 m/s, angle 25.7掳 above horizontal.

Step by step solution

01

Understand the Problem

The ski jumper lands 51.0 m away horizontally, his final velocity is 23.0 m/s at an angle of 43.0掳 below the horizontal. We need to find his initial velocity and its angle relative to the horizontal when he left the ramp.
02

Break Down the Final Velocity

Since the final velocity is given as 23.0 m/s at an angle 43.0掳 below horizontal, we break it into components. The horizontal component is \(v_{fx} = 23.0 \cos(43.0^\circ) \) and the vertical component is \(v_{fy} = -23.0 \sin(43.0^\circ)\) (negative because it is downward).
03

Determine Time of Flight

Since the horizontal velocity \(v_{fx}\) remains constant (no air resistance), the time of flight \( t \) is given by \( t = \frac{51.0}{v_{fx}}\).
04

Calculate Initial Vertical Velocity

Use the kinematic equation for vertical motion: \(v_{fy} = v_{iy} - gt\). Solve for \(v_{iy} = v_{fy} + gt\) where \(g\) is 9.81 m/s虏.
05

Find Initial Horizontal Velocity

Since there is no horizontal acceleration, the initial horizontal velocity component \(v_{ix}\) is the same as \(v_{fx}\).
06

Calculate Initial Velocity Magnitude

Use Pythagorean theorem: the magnitude of initial velocity \(v_i\) is \(v_i = \sqrt{v_{ix}^2 + v_{iy}^2}\).
07

Determine Initial Velocity Angle

Find the angle \(\theta\) using \(\tan(\theta) = \frac{v_{iy}}{v_{ix}}\) and solve for \(\theta = \tan^{-1}\left(\frac{v_{iy}}{v_{ix}}\right)\).

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Key Concepts

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

Kinematic Equations
When analyzing projectile motion, kinematic equations serve as our foundation. They help us understand how different parameters such as velocity, time, and position interrelate over time. These equations are particularly critical when the object in motion is affected by gravity.

The equations relevant for projectile motion include:
  • Vertical motion: \[v_{fy} = v_{iy} - gt\]
  • Horizontal motion: \[x = v_{ix} \cdot t\]
  • The equation linking velocity and acceleration: \[v^2 = v_{i}^2 + 2a \cdot d\]
In these equations, vertical and horizontal motions are treated separately because gravity only affects vertical motion. The equations allow us to calculate unknowns by using known values, making them essential for solving projectile problems like the ski jumper's leap.
Velocity Components
In projectile motion, breaking down velocity into components allows us to analyze different forces acting in various directions.

The velocity of the ski jumper before landing, angled below the horizontal, needs to be split into horizontal and vertical components. By using trigonometric functions:
  • Horizontal component: \[v_{fx} = v \cdot \cos(43.0^\circ)\]
  • Vertical component: \[v_{fy} = -v \cdot \sin(43.0^\circ)\] (negative due to downward direction)
This breakdown allows us to treat the horizontal and vertical paths separately and determine the time of flight and initial conditions.
Time of Flight
Time of flight is a crucial aspect of any projectile motion problem, including our ski jumper. It is the time duration for which the object remains in the air.

Since there is no air resistance, the horizontal velocity stays constant. Therefore, we calculate the time of flight by using the horizontal component of velocity:
  • \[t = \frac{51.0 \, \text{m}}{v_{fx}}\]
The horizontal displacement is known, and this relation helps us find how long the ski jumper was in the air. With the time of flight known, we can further calculate the initial vertical velocity component.
Initial Velocity Calculation
Finding the initial velocity of the ski jumper involves piecing together the components we've calculated. We already know:
  • The initial horizontal velocity component \(v_{ix}\) is derived directly from the horizontal component of the final velocity.
  • Using time of flight and kinematic equations, the initial vertical velocity \(v_{iy}\) can be found: \[v_{iy} = v_{fy} + gt\]
Once we have these components, we calculate the total initial velocity's magnitude with the Pythagorean theorem:
  • \[v_i = \sqrt{v_{ix}^2 + v_{iy}^2}\]
Additionally, the initial velocity angle relative to the horizontal is found using:
  • \[\theta = \tan^{-1}\left(\frac{v_{iy}}{v_{ix}}\right)\]
These calculations help us fully understand the ski jumper's motion right from the ramp's end.

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Most popular questions from this chapter

You are in a hot-air balloon that, relative to the ground, has a velocity of \(6.0 \mathrm{m} / \mathrm{s}\) in a direction due east. You see a hawk moving directly away from the balloon in a direction due north. The speed of the hawk relative to you is \(2.0 \mathrm{m} / \mathrm{s}\). What are the magnitude and direction of the hawk's velocity relative to the ground? Express the directional angle relative to due east.

A Coast Guard ship is traveling at a constant velocity of \(4.20 \mathrm{m} / \mathrm{s}\), due east, relative to the water. On his radar screen the navigator detects an object that is moving at a constant velocity. The object is located at a distance of \(2310 \mathrm{m}\) with respect to the ship, in a direction \(32.0^{\circ}\) south of east. Six minutes later, he notes that the object's position relative to the ship has changed to \(1120 \mathrm{m}, 57.0^{\circ}\) south of west. What are the magnitude and direction of the velocity of the object relative to the water? Express the direction as an angle with respect to due west.

A soccer player kicks the ball toward a goal that is \(16.8 \mathrm{m}\) in front of him. The ball leaves his foot at a speed of \(16.0 \mathrm{m} / \mathrm{s}\) and an angle of \(28.0^{\circ}\) above the ground. Find the speed of the ball when the goalie catches it in front of the net.

The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his head, forcing him to move quickly away from the net (see the drawing). Suppose that you lob the ball with an initial speed of \(15.0 \mathrm{m} / \mathrm{s},\) at an angle of \(50.0^{\circ}\) above the horizontal. At this instant your opponent is \(10.0 \mathrm{m}\) away from the ball. He begins moving away from you 0.30 s later, hoping to reach the ball and hit it back at the moment that it is \(2.10 \mathrm{m}\) above its launch point. With what minimum average speed must he move? (Ignore the fact that he can stretch, so that his racket can reach the ball before he does.

Two boats are heading away from shore. Boat 1 heads due north at a speed of \(3.00 \mathrm{m} / \mathrm{s}\) relative to the shore. Relative to boat \(1,\) boat 2 is moving \(30.0^{\circ}\) north of east at a speed of \(1.60 \mathrm{m} / \mathrm{s} .\) A passenger on boat 2 walks due east across the deck at a speed of \(1.20 \mathrm{m} / \mathrm{s}\) relative to boat 2 What is the speed of the passenger relative to the shore?
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