Question

An object is thrown vertically and has an upward velocity of \(25 \mathrm{~m} / \mathrm{s}\) when it reaches one fourth of its maximum height above its launch point. What is the initial (launch) speed of the object?

Short answer

Solution: 1. The final velocity at maximum height is 0 m/s. 2. Calculate the time taken to reach one-fourth of the maximum height using the formula: t = (u - v) / g. 3. Calculate the distance traveled to reach one-fourth of the maximum height using the formula: s = ut - (1/2)gt^2. 4. Determine the initial velocity by using the equations: H = u^2 / 2g and s = u^2 / 8g. After solving for the initial launch speed, u, we'll find the answer to the question.

Step-by-step solution

Calculate the final velocity at maximum height

At maximum height, the final vertical velocity of the object is 0. We will use this fact in further calculations.

Calculate the time taken to reach one-fourth of the maximum height

Let's use the equation of motion: \(v = u - gt\), where \(v\) is the final velocity, \(u\) is the initial velocity, \(g\) is the acceleration due to gravity (\(9.8 \mathrm{~m} / \mathrm{s^2}\)), and \(t\) is the time. Rearranging the equation for \(t\), we have: \(t = \frac{u - v}{g}\). We will use this equation to determine the time taken to reach one-fourth of the maximum height.

Calculate the distance traveled to reach one-fourth of the maximum height

We can use the equation of motion: \(s = ut - \frac{1}{2}gt^2\), where \(s\) is the distance traveled. Substituting the known values of \(v\), \(g\), and \(t\) from Step 1 and Step 2, we can calculate the distance traveled to reach one-fourth of the maximum height.

Determine the initial velocity

Now, let's express the maximum height, \(H\), that the object reaches in terms of the initial velocity, \(u\), and the acceleration due to gravity, \(g\). Using the formula \(v^2 = u^2 - 2gH\), we have \(H = \frac{u^2}{2g}\). According to the given problem, the object is at one-fourth its maximum height when its upward velocity is \(25\mathrm{~m} / \mathrm{s}\). So, the distance traveled, \(s\), is equal to: \(s = \frac{1}{4}H = \frac{u^2}{8g}\). Now, we plug the values from Step 3 into this equation and solve for \(u\). After calculating, we'll get the initial launch speed of the object.

Key concepts

Kinematics Equations

Kinematics equations, also known as the equations of motion, are fundamental to understanding motion in physics. They describe the relation between displacement, velocity, acceleration, and time without taking the cause of motion into account. One primary equation is: \[ v = u + at \]where \(v\) is final velocity, \(u\) is initial velocity, \(a\) is acceleration, and \(t\) is time.
Another important equation is:\[ s = ut + \frac{1}{2}at^2 \]where \(s\) is the displacement.
These equations can be modified to eliminate the variable of time, resulting in:\[ v^2 = u^2 + 2as \]This particular equation is useful when calculating aspects of projectile motion, such as the initial launch speed of an object thrown vertically.

Projectile Motion

Projectile motion refers to the motion of an object that is thrown or projected into the air and is subject only to the acceleration due to gravity. The path of a projectile is called its trajectory, which is typically a parabolic shape due to the constant influence of gravity. When an object is launched with an initial speed and angle, it follows this type of motion.
For vertical projectile motion, there are key moments to consider: the launch, the peak height where the velocity becomes zero, and the descent. The initial launch speed significantly affects the maximum height reached and the total time the object will spend in the air. Analyzing these stages with kinematics equations enables us to find unknown variables, such as the initial launch speed.

Acceleration Due to Gravity

Acceleration due to gravity is the rate at which an object accelerates as it falls toward Earth, due to Earth's gravitational pull. This acceleration is denoted by \(g\), and on Earth, it has an average value of about \(9.8 \text{ m/s}^2\).
In physics problems, including projectile motion, \(g\) is always directed downwards towards the center of the Earth. It's important to remember that this acceleration remains constant for any object in free fall near the Earth's surface, regardless of its mass. When calculating the initial launch speed of an object in projectile motion, \(g\) helps determine how quickly the object slows down as it ascends and accelerates as it descends.

Motion Under Gravity

Motion under gravity is a term used to describe the movement of an object solely under the influence of gravitational acceleration. This type of motion is often seen in free-falling objects and vertical projectiles.
An object thrown vertically into the air will rise until gravity stops its ascent, then it will start its descent back to the original launch point. The kinematics equations applied to motion under gravity allow us to calculate various aspects such as the time taken to reach a certain height, the speed at different points during the object's flight, and, crucially for our textbook problem, the initial launch speed. These calculations are essential in predicting how objects will behave when launched or dropped in Earth's gravitational field.