Question

A skier launches off a ski jump with a horizontal velocity of \(30.0 \mathrm{~m} / \mathrm{s}\) (and no vertical velocity component). What are the magnitudes of the horizontal and vertical components of her velocity the instant before she lands 2.00 s later?

Short answer

Answer: The magnitudes of the horizontal and vertical components of the skier's velocity just before landing are: - Horizontal component: \(30.0\mathrm{~m}/\mathrm{s}\) - Vertical component: \(19.65\mathrm{~m}/\mathrm{s}\)

Step-by-step solution

Identify given values

The given values in the problem are: - Initial horizontal velocity (\(v_{0x}\)) = \(30.0\mathrm{~m} / \mathrm{s}\) - Initial vertical velocity (\(v_{0y}\)) = \(0\mathrm{~m} / \mathrm{s}\) (since no vertical velocity component is mentioned) - Time of flight (t) = \(2.00\mathrm{~s}\)

Calculate the vertical displacement

To find the final vertical velocity component, we first need to determine the vertical displacement (Δy) during the time of flight. We'll use the following equation of motion for this: Δy = \(v_{0y}t + \frac{1}{2}gt^2\) where: - Δy is the vertical displacement - \(v_{0y}\) is the initial vertical velocity (which is 0 in this case) - t is the time of flight - g is the acceleration due to gravity, approximately \(-9.81\mathrm{~m}/\mathrm{s}^2\) So the vertical displacement is: Δy = \((0)(2.00) + \frac{1}{2}(-9.81)(2.00)^2\) Δy = \(-19.62\mathrm{~m}\)

Calculate the final vertical velocity component

Now let's find the final vertical velocity component (v_y) using the following equation: \(v_{y}^2 = v_{0y}^2 + 2gΔy\) where: - \(v_{y}\) is the final vertical velocity component - \(v_{0y}\) is the initial vertical velocity (which is 0 in this case) - g is the acceleration due to gravity, approximately \(-9.81\mathrm{~m}/\mathrm{s}^2\) - Δy is the vertical displacement So the final vertical velocity component is: \(v_{y}^2 = (0)^2 + 2(-9.81)(-19.62)\) \(v_{y}^2 = 386.0844\) Taking the square root of both sides, we get: \(v_{y} = \pm 19.65\mathrm{~m}/\mathrm{s}\) Since the skier is landing, the vertical velocity component is directed downwards, so we choose the negative value: \(v_{y} = -19.65\mathrm{~m}/\mathrm{s}\)

Determine the horizontal velocity component

Since there is no horizontal acceleration during the skier's flight, the horizontal velocity component remains constant throughout: \(v_{x} = v_{0x} = 30.0\mathrm{~m}/\mathrm{s}\)

State the magnitudes of the horizontal and vertical components

The magnitudes of the horizontal and vertical components of the skier's velocity just before landing are: - Horizontal component: \(v_{x} = 30.0\mathrm{~m}/\mathrm{s}\) - Vertical component: \(v_{y} = 19.65\mathrm{~m}/\mathrm{s}\)

Key concepts

Horizontal Velocity Component

The horizontal velocity component refers to the speed at which an object moves along the horizontal axis. In a projectile motion scenario like the exercise with the skier, the horizontal velocity component, often denoted as \( v_{x} \), remains constant if there's no horizontal force acting upon the object.

In the case of our skier, who launches with an initial horizontal velocity of \( 30.0 \mathrm{~m} / \mathrm{s} \) and experiences no air resistance or any other form of horizontal force, this initial velocity will remain the same throughout the flight. That's why, even just before landing, the skier's horizontal velocity component is still \( 30.0 \mathrm{~m} / \mathrm{s} \).

Vertical Velocity Component

On the other hand, the vertical velocity component (\( v_{y} \)) is influenced by gravity and changes over time. An object thrown into the air with an initial vertical velocity will slow down on the way up, stop momentarily at the peak, then speed up again as it falls back down, all due to the constant acceleration of gravity pulling it toward the Earth.

In our exercise, since the skier lacks an initial vertical component, \( v_{0y} \) is zero. However, she doesn't stay at zero vertical velocity because gravity affects her motion the moment she is airborne, accelerating her downward. We use the equation \( v_{y} = v_{0y} + g \cdot t \) to describe how gravity changes her vertical velocity over time; in this context, we use the square root of the equation \( v_{y}^2 = v_{0y}^2 + 2g\Delta y \) to find her velocity just before impact.

Equations of Motion

Kinematics involves several key equations to describe the motion of objects. For vertically launched projectiles, we often use \( \Delta y = v_{0y}t + \frac{1}{2}g t^2 \) to find the displacement and \( v_y^2 = v_{0y}^2 + 2g \Delta y \) to find the final vertical velocity.

These equations stem from integrating the acceleration due to gravity over time to get velocity, and integrating velocity over time to get displacement. They allow us to predict where and how fast the object will be at any given point during its motion. Understanding these equations is crucial for solving problems involving projectile motion, as seen in the step-by-step solution to the exercise.

Acceleration Due to Gravity

The acceleration due to gravity, denoted by the symbol \( g \), is approximately \( 9.81 \, \mathrm{m/s^2} \) on the surface of the Earth. This acceleration is the rate at which an object speeds up as it falls, assuming no other forces act on it, such as air resistance.

Gravity is a constant acceleration affecting all objects equally, pulling them towards the center of the Earth. Whether a feather or a hammer is dropped, they would theoretically accelerate at the same rate in a vacuum. In our skier's case, gravity is what causes her vertical velocity component to increase in magnitude in the downward direction over the duration of her jump.