Question

A rock is tossed off the top of a cliff of height \(34.9 \mathrm{~m}\) Its initial speed is \(29.3 \mathrm{~m} / \mathrm{s}\), and the launch angle is \(29.9^{\circ}\) with respect to the horizontal. What is the speed with which the rock hits the ground at the bottom of the cliff?

Short answer

Answer: The speed of the rock when it hits the ground is approximately 33.66 m/s.

Step-by-step solution

Identify known values and the desired output

We are given the following information: - Height of the cliff: \(h = 34.9 \mathrm{~m}\) - Initial speed: \(v_0 = 29.3 \mathrm{~m/s}\) - Launch angle: \(\theta = 29.9^{\circ}\) - Desired output: The speed of the rock when it hits the ground

Determine the initial velocity components

We can separate the initial velocity into horizontal (\(v_{0x}\)) and vertical (\(v_{0y}\)) components using the launch angle: - \(v_{0x} = v_0 \cdot \cos{\theta} = 29.3 \mathrm{~m/s} \cdot \cos{29.9^{\circ}} \approx 25.54 \mathrm{~m/s}\) - \(v_{0y} = v_0 \cdot \sin{\theta} = 29.3 \mathrm{~m/s} \cdot \sin{29.9^{\circ}} \approx 14.63 \mathrm{~m/s}\)

Calculate the time it takes for the rock to reach the ground

First, we need to find the time it takes for the rock to reach the ground. In the vertical direction, the time it takes to hit the ground can be found using the following equation (assuming constant acceleration due to gravity \(g = 9.81 \mathrm{~m/s^2}\)): $$ h = v_{0y}\cdot t - \dfrac{1}{2} \cdot g \cdot t^2$$ Solving for \(t\), we have: $$ t = \dfrac{v_{0y} + \sqrt{v_{0y}^2 + 2\cdot h \cdot g}}{g} = \dfrac{14.63 \mathrm{~m/s} + \sqrt{(14.63 \mathrm{~m/s})^2 + 2 \cdot 34.9 \mathrm{~m} \cdot 9.81 \mathrm{~m/s^2}}}{9.81 \mathrm{~m/s^2}} \approx 3.76 \mathrm{~s}$$

Calculate the final velocity components

As the horizontal component of velocity does not change (assuming no air resistance), \(v_{fx} = v_{0x} \approx 25.54 \mathrm{~m/s}\) We can now find the final vertical component (\(v_{fy}\)) using the following equation and the calculated time: $$ v_{fy} = v_{0y} - g\cdot t \approx 14.63 \mathrm{~m/s} - 9.81 \mathrm{~m/s^2} \cdot 3.76 \mathrm{~s} \approx -21.65 \mathrm{~m/s}$$

Find the final speed of the rock

To find the final speed of the rock, we can combine the final horizontal and vertical components using the Pythagorean theorem: $$ v_f = \sqrt{v_{fx}^2 + v_{fy}^2} = \sqrt{(25.54 \mathrm{~m/s})^2 + (-21.65 \mathrm{~m/s})^2} \approx 33.66 \mathrm{~m/s}$$ Therefore, the speed with which the rock hits the ground at the bottom of the cliff is approximately \(33.66 \mathrm{~m/s}\).

Key concepts

Kinematics

Kinematics is a branch of physics that focuses on the motion of objects without considering the forces that cause this motion. It's like telling a story of how something is moving; where and when it's going, without delving into why it's moving that way. In projectile motion, which is a common kinematic scenario, an object moves in a two-dimensional space subject to Earth's gravitational pull.

Here's how it unfolds: Imagine tossing a rock off a cliff. Once it leaves your hand, it follows a parabolic path until it hits the ground. Kinematics allows us to describe this path using variables such as the initial velocity, its components, the launch angle, and the acceleration due to gravity. These variables are key for understanding exactly how the rock will move through the air. We use equations of motion to predict where the rock will be at any given time and at what speed it will smash into the ground.

Initial Velocity Components

The concept of initial velocity components is central to solving projectile motion problems. At the moment of launch, the rock's initial velocity can be split into two perpendicular components: horizontal and vertical. This is like slicing a pie into two pieces right at the launch angle. The horizontal component, which we denote as v_0x, doesn't change much during the flight because in our idealized scenario, we ignore things like air resistance.

To find this horizontal component, you take the initial speed and multiply it by the cosine of the launch angle: v_0x = v₀cos(θ). Similarly, for the vertical bit, that's the slice going up and down, we multiply by the sine instead: v_0y = v₀sin(θ). This vertical slice is going to change as the rock goes up and comes back down, due to gravity pulling it back towards the Earth.

Acceleration Due to Gravity

Gravity is like the invisible hand that shapes the trajectory of a tossed rock. On Earth, this force of attraction provides a constant acceleration towards the ground, designated as g, approximately 9.81 m/s². Whether an object is moving up or down, gravity will always accelerate it downwards. That's why the rock, after being tossed, eventually falls down the cliff no matter how hard or at what angle you throw it.

When we calculate the rock’s motion, this constant acceleration due to gravity comes into play. For projectile motion, it doesn't accelerate the horizontal motion - only vertical. Hence, as the rock falls, its vertical speed increases by about 9.81 m/s each second, which dramatically affects its path and the impact speed when it finally collides with the ground.