Question

A girl is standing at the edge of a cliff \(100 . \mathrm{m}\) above the ground. She reaches out over the edge of the cliff and throws a rock straight upward with a speed \(8.00 \mathrm{~m} / \mathrm{s}\). a) How long does it take the rock to hit the ground? b) What is the speed of the rock the instant before it hits the ground?

Short answer

Question: A girl throws a rock vertically upward from the edge of a cliff. The rock reaches a height of 100 m above its initial position, and the initial speed of the rock is 8 m/s. Calculate the total time it takes for the rock to hit the ground and the speed of the rock right before it reaches the ground. Answer: The total time it takes for the rock to hit the ground is approximately 5.403 seconds, and the speed of the rock right before it hits the ground is about -44.998 m/s.

Step-by-step solution

a) Finding the time for the rock to hit the ground

We can split the motion into two parts: the upward phase and the downward phase. First, let's find the time it takes for the rock to reach its highest point. When the rock reaches its highest point, its velocity will be 0. We can use the equation: \(v = u - gt\) where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity (9.81 m/s²) and t is the time. Now, let's solve for t: \(0 = 8.00 - 9.81t\) \(t = \frac{8.00}{9.81} \approx 0.816 s\) Now we have to find the time it takes for the rock to fall from the highest point to the ground. Since we know the upward phase height is the same as the downward phase height, we can use the equation: \(h = \frac{1}{2}gt^2\) Here, h is the height reached from the initial position and g is the acceleration due to gravity (9.81 m/s²). We already know that the total height is 100 m, so we need to add the height reached during the upward phase to the overall height. Total height = 100 + h \(h = \frac{1}{2}(9.81)(0.816)^2 \approx 3.25 m\) Total height = 103.25 m Now, we can solve for the time it takes for the rock to fall from its highest point to the ground using the same equation we used earlier: \(103.25 = \frac{1}{2}(9.81)t^2\) \(t^2 = \frac{103.25}{4.905} \approx 21.052\) \(t \approx \sqrt{21.052} \approx 4.587 s\) Finally, we can find the total time by adding the time taken during the upward phase and the downward phase: Total time = 0.816 + 4.587 ≈ 5.403 s

b) Finding the speed of the rock right before it hits the ground

We can use the equation: \(v^2 = u^2 + 2gh\) where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity (9.81 m/s²), and h is the height. Since the rock falls freely from its highest point, the initial velocity(u) will be 0. Now we have: \(v^2 = 0 + 2(9.81)(103.25)\) \(v^2 \approx 2024.795\) \(v \approx \sqrt{2024.795} \approx 44.998 m/s\) Since the rock is falling downward, the speed will be negative: Final speed = -44.998 m/s Therefore, the time it takes the rock to hit the ground is approximately 5.403 seconds, and the speed of the rock right before it hits the ground is about -44.998 m/s.

Key concepts

Kinematics

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. It focuses on the concepts of velocity, acceleration, displacement, and time. These parameters can describe the motion in one, two, or three dimensions. In the provided exercise, kinematics is applied to a two-dimensional scenario, including the effects of gravity, and analyzes the motion of a rock thrown vertically from the edge of a cliff. The key to understanding kinematics is to break down the motion into understandable parts. The upward motion until the rock stops, and the subsequent fall, are considered separately which simplifies the analysis and calculations.

Key aspects of kinematics in projectile motion include:

• The initial velocity given to the projectile, in this case, the rock.
• The time it takes to reach the peak of its motion where its velocity reduces to zero.
• The height and time taken for the fall after reaching the peak.

Understanding the relationship between these aspects is crucial for solving any kinematic problem, as they are fundamental for the application of motion equations.

Free-fall Acceleration

Free-fall acceleration is a specific type of acceleration experienced by an object that is only under the influence of gravity. The standard value of free-fall acceleration on Earth's surface is approximately 9.81 meters per second squared (\(9.81 \text{ m/s}^2\)). It's important to note that this value can vary slightly depending on altitude and location but is often used as a constant in physics problems involving free-fall.

When an object is thrown upwards, it slows down due to the force of gravity pulling it back to Earth. Upon reaching the top of its path or the peak of its trajectory, its velocity momentarily becomes zero, and then gravity accelerates it on its way back down. Gravity affects all objects equally, regardless of their mass, and this constant acceleration influences the calculation of various motion parameters such as displacement, velocity, and time.

In the solution provided for the exercise, the acceleration due to gravity plays a key role in determining both the time it takes for the rock to reach its highest point and to fall back down to the ground, as well as the final velocity before impact.

Motion Equations

Motion equations, also known as the equations of kinematics, describe the relationships between velocity, acceleration, time, and displacement. They allow us to predict future motion or reconstruct the past motion of an object. The basic motion equations applicable to uniformly accelerated motion, like free-fall, include:

• \(v = u + at\) - which relates velocities, acceleration, and time.
• \(s = ut + \frac{1}{2}at^2\) - which connects displacement, initial velocity, acceleration, and time.
• \(v^2 = u^2 + 2as\) - which links the final velocity, initial velocity, acceleration, and displacement.

In our exercise, the initial upward throw and the subsequent free-fall can be described using these equations. To solve part (a), we used the first and second equations to find the time the rock takes to reach the highest point and to fall from there to the ground. In part (b), the third equation helped us find the final velocity just before impact.

These motion equations are extremely powerful as they can be used to solve for one unknown variable when the other variables are known, making them essential tools for answering many kinematics-related problems.