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A ball is thrown toward a cliff of height \(h\) with a speed of \(30 \mathrm{m} / \mathrm{s}\) and an angle of \(60^{\circ}\) above horizontal. It lands on the edge of the cliff 4.0 s later. a. How high is the cliff? b. What was the maximum height of the ball? c. What is the ball's impact speed?

Short Answer

Expert verified
a. The cliff is 53.92 m high. b. The maximum height of the ball is 88.22 m. c. The ball's impact speed is 20.54 m/s.

Step by step solution

01

Determine the vertical and horizontal components of initial velocity

First, resolve the velocity into its components. The horizontal velocity \(v_x\) is given by \(30 \, cos(60^{\circ}) = 15 \, m/s\). The vertical component of velocity \(v_{0y}\) is given by \(30 \, sin (60^{\circ}) = 25.98 \, m/s\). The speed of 30 m/s was resolved using sine and cosine of the launch angle.
02

Calculate the height of the cliff

The height 'h' from the point of launch to the edge of the cliff after 4.0s is calculated using the equation of motion as: \(h = v_{0y}t - 0.5gt^2\), where \(g = 9.8 \, m/s^2\) is the acceleration due to gravity and \(t = 4.0 \, s\). Substituting the values of \(v_{0y}\), \(g\), \(t\) in the equation the height 'h' can be found as \(h = (25.98 \, m/s)(4.0 \, s) - 0.5(9.8 \, m/s^2)(4.0 \, s)^2 = 53.92 \, m\).
03

Find the maximum height of the ball

When the ball reaches the maximum height, the vertical velocity becomes zero. The time to reach this point can be calculated using the equation \(v_{fy} = v_{0y} - gt\), giving \(t = v_{0y}/g = 25.98 \, m/s \div 9.8 \, m/s^2 \approx 2.65 \, s\). The max height 'h\_max' at this time is then given by equation \(h\_max = v_{0y}t - 0.5gt^2\), resulting in \( h\_max = 25.98 \, m/s \times 2.65 \, s - 0.5 \times 9.8 \, m/s^2 \times (2.65 \, s)^2 \approx 34.3 \, m\). Note that this height is calculated from the launch point, we need to add the cliff height to get the absolute max height, so the maximum height of the ball 'H\_max' is \(H\_max = h + h\_max = 53.92 \, m + 34.3 \, m = 88.22 \, m\).
04

Determine the ball's impact speed

The ball's impact speed is the magnitude of the resultant of the final horizontal and vertical velocities. The horizontal velocity remains constant throughout the journey, i.e., \(v_{fx} = v_x = 15 \, m/s\), and the final vertical velocity \(v_{fy}\) can be calculated using the equation \(v_{fy} = v_{0y} - gt\), giving \(v_{fy} = 25.98 \, m/s - 9.8 \, m/s^2 \times 4.0 \, s = -14.02 \, m/s\). The impact speed \(v_f\) is then the magnitude of the final velocity vector, given by \(\sqrt{(v_{fx})^2 + (v_{fy})^2} = \sqrt{(15 \, m/s)^2 + (-14.02 \, m/s)^2} \approx 20.54 \, m/s\).

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Key Concepts

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

Kinematics Equations
Kinematics equations are the foundation of analyzing motion in physics. They describe the relationships between displacement, velocity, acceleration, and time without requiring information about the forces that cause such motion. For projectile motion, there are two key components: vertical and horizontal motion. The vertical component is influenced by gravity, while the horizontal motion happens at a constant velocity because it is unaffected by gravity in the absence of air resistance.

To solve problems involving kinematics, one typically uses a set of four equations:
  • \( v = u + at \) where 'v' is the final velocity, 'u' is the initial velocity, 'a' is the acceleration, and 't' is the time taken.
  • \( s = ut + \frac{1}{2}at^2 \) where 's' is the displacement.
  • \( v^2 = u^2 + 2as \) relates the velocities to displacement and acceleration.
  • \( s = vt - \frac{1}{2}at^2 \) another form of the displacement equation.
For the projectile motion problems, gravity (\( -9.8 \text{ m/s}^2 \) is used as 'a' in the vertical component, and the horizontal acceleration is zero.
Trajectory Calculation
Calculating the trajectory of a projectile involves understanding its path through space as a function of time. For an object launched at an angle, its trajectory is parabolic due to the constant acceleration of gravity acting downward. To analyze the motion, we separate it into two components: horizontal (\( x \) direction) and vertical (\( y \) direction).

These components are independent, with the horizontal motion at a constant speed and the vertical motion being affected by gravity. By determining the time of flight and the maximum height, the trajectory calculation gives us the overall path of the projectile. The key is to realize that motion in the 'x' direction doesn't affect motion in the 'y' direction and vice versa.
Physics Problem Solving
When solving a physics problem, understanding the concept is as crucial as applying the formulas correctly. A systematic approach is recommended:
  1. Identify the known and unknown variables.
  2. Visualize the problem by sketching a diagram.
  3. Decide which principles of physics are applicable.
  4. Break down the problem into manageable parts if necessary.
  5. Solve algebraically, keeping track of units and significant figures.
  6. Check if the answers are reasonable and consistent with the problem.
For projectile motion, clarity on the components of motion, the effect of gravity, and the kinematic equations are vital. By practicing these steps methodically, students enhance their understanding and problem-solving skills.
Initial Velocity Components
Initial velocity components are vital to projectile motion, as they set the stage for the entire trajectory. When an object is launched at an angle, its initial velocity has both horizontal (\( v_x \) and vertical (\( v_y \) components, defined by:
  • \( v_x = v \times \text{cos}(\theta) \) - the horizontal component,
  • \( v_y = v \times \text{sin}(\theta) \) - the vertical component,
where '\theta' is the launch angle and 'v' is the launch speed. The initial horizontal velocity remains constant throughout the flight, and the initial vertical velocity is used to calculate maximum height, flight time, and other important aspects of motion. For the given problem, a 30 m/s launch speed at a 60-degree angle yields initial velocity components needed to calculate various stages of the projectile's flight.

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