Projectile Motion and Relative Motion: Step-by-Step Physics Guidance

Q1. Firemen are shooting a stream of water at a burning building using a high-pressure hose that shoots out the water with a speed of 25.0 m/s as it leaves the end of the hose. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation α of the hose until the water takes 3.00 s to reach a building 45.0 m away. You can ignore air resistance; assume that the end of the hose is at ground level.

• (a) Find the angle of elevation α.

• (b) Find the speed and acceleration of the water at the highest point in its trajectory.

• (c) How high above the ground does the water strike the building, and how fast is it moving just before it hits the building?

Background

Topic: Projectile Motion

This problem tests your understanding of two-dimensional projectile motion, including how to resolve initial velocity into components, use kinematic equations, and analyze motion at specific points in the trajectory.

Key Terms and Formulas

• Projectile motion: The motion of an object thrown or projected into the air, subject only to acceleration due to gravity.

• Initial velocity components:

  • $v_{0x} = v_0 \cos \alpha$

  • $v_{0y} = v_0 \sin \alpha$

• Horizontal displacement: $x = v_{0x} t$

• Vertical displacement: $y = v_{0y} t - \frac{1}{2} g t^2$

• Acceleration due to gravity: $g = 9.8\ \mathrm{m/s^2}$

Step-by-Step Guidance

1. Start by identifying the known values: $v_0 = 25.0\ \mathrm{m/s}$, $t = 3.00\ \mathrm{s}$, $x = 45.0\ \mathrm{m}$, $y_0 = 0$ (ground level).

2. Write the equations for horizontal and vertical motion. For horizontal motion: $x = v_0 \cos \alpha \cdot t$.

3. Rearrange the horizontal equation to solve for $\cos \alpha$:

   $\cos \alpha = \frac{x}{v_0 t}$

   For vertical motion, use $y = v_0 \sin \alpha \cdot t - \frac{1}{2} g t^2$ to find the height when the water reaches the building.

4. To find the speed and acceleration at the highest point, recall that at the peak, the vertical velocity component is zero, but the horizontal component remains unchanged. The acceleration is always $g$ downward.

Try solving on your own before revealing the answer!

Final Answers:

• (a) $\alpha \approx 36.9^\circ$

• (b) Speed at the highest point: $20.0\ \mathrm{m/s}$; Acceleration: $9.8\ \mathrm{m/s^2}$ downward

• (c) Height above ground: $7.5\ \mathrm{m}$; Speed just before hitting: $25.0\ \mathrm{m/s}$

Each part uses the kinematic equations for projectile motion, with careful attention to the components of velocity and the effects of gravity.

Q2. A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/s. A 1.0-kg stone is thrown from the basket with an initial velocity of 15.0 m/s perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. The person in the basket sees the stone hit the ground 6.00 s after being thrown. Assume that the balloon continues its downward descent with the same constant speed of 20.0 m/s.

• (a) How high was the balloon when the rock was thrown out?

• (b) How high is the balloon when the rock hits the ground?

• (c) At the instant the rock hits the ground, how far is it from the basket?

• (d) Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer (i) at rest in the basket and (ii) at rest on the ground.

Background

Topic: Relative Motion and Two-Dimensional Kinematics

This problem involves analyzing motion from different reference frames, using kinematic equations for both vertical and horizontal motion, and understanding how velocities add in different frames.

Key Terms and Formulas

• Relative velocity: $\vec{v}_{\text{stone, ground}} = \vec{v}_{\text{stone, basket}} + \vec{v}_{\text{basket, ground}}$

• Vertical displacement: $y = v_{0y} t + \frac{1}{2} a t^2$

• Horizontal displacement: $x = v_{0x} t$

• Acceleration due to gravity: $g = 9.8\ \mathrm{m/s^2}$

Step-by-Step Guidance

1. For part (a), set up the vertical motion equation for the stone, considering its initial vertical velocity (same as the balloon's downward velocity) and the time until it hits the ground.

   $y = v_{0y} t + \frac{1}{2} (-g) t^2$

2. Plug in $v_{0y} = -20.0\ \mathrm{m/s}$ (downward), $t = 6.00\ \mathrm{s}$, and $g = 9.8\ \mathrm{m/s^2}$ to solve for the initial height $y$.

3. For part (b), calculate how far the balloon descends in 6.00 s using $\Delta y = v_{\text{balloon}} \cdot t$.

4. For part (c), determine the horizontal distance the stone travels in 6.00 s using $x = v_{0x} t$, where $v_{0x} = 15.0\ \mathrm{m/s}$ (perpendicular to descent).

5. For part (d), find the stone's velocity components just before it hits the ground, both relative to the basket and to the ground. For the ground frame, add the balloon's velocity to the stone's initial velocity components.

Try solving on your own before revealing the answer!

Final Answers:

• (a) $y_0 = 196\ \mathrm{m}$

• (b) $y = 76\ \mathrm{m}$

• (c) $x = 90.0\ \mathrm{m}$

• (d) (i) Relative to basket: $v_x = 15.0\ \mathrm{m/s}$, $v_y = -78.8\ \mathrm{m/s}$; (ii) Relative to ground: $v_x = 15.0\ \mathrm{m/s}$, $v_y = -98.8\ \mathrm{m/s}$

Each part uses kinematic equations and the concept of relative velocity to analyze the motion from different perspectives.

Q3. In a physics lab, you attach a 0.200-kg air-track glider to the end of an ideal spring of negligible mass and start it oscillating.

Background

Topic: Simple Harmonic Motion (SHM)

This question is about the motion of a mass-spring system, which is a classic example of simple harmonic motion. You may be asked to analyze the period, frequency, amplitude, or energy of the oscillating system.

Key Terms and Formulas

• Simple harmonic motion: Oscillatory motion under a restoring force proportional to displacement.

• Period of oscillation: $T = 2\pi \sqrt{\frac{m}{k}}$

• Frequency: $f = \frac{1}{T}$

• Maximum speed: $v_{\text{max}} = \omega A$ where $\omega = 2\pi f$

• Total energy: $E = \frac{1}{2} k A^2$

Step-by-Step Guidance

1. Identify the mass $m = 0.200\ \mathrm{kg}$ and note that the spring is ideal (no mass, no damping).

2. Recall the formula for the period of a mass-spring system: $T = 2\pi \sqrt{\frac{m}{k}}$.

3. If the spring constant $k$ or amplitude $A$ is given, use the relevant formulas for frequency, maximum speed, or energy as needed.

4. Set up the equations for the specific quantity you are asked to find (e.g., period, frequency, energy), but do not solve for the final value yet.

Try solving on your own before revealing the answer!

Final Answer:

The period, frequency, and other properties can be found using the above formulas once the spring constant or amplitude is provided.