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Chapter 2: Problem 71

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

Expert verified
Answer: The initial speed of the object is approximately 35.36 m/s.

Step by step solution

01

Write down the given information

We are given that - The object's upward velocity is 25 m/s when it reaches 1/4 of its maximum height - We need to find the initial (launch) speed of the object
02

Apply the Kinematic Equations

To solve this problem, we can use the following kinematic equation for vertical motion: $$ v^2 = u^2 + 2as $$ where - \(v\) is the final velocity, which is given at 25 m/s - \(u\) is the initial velocity that we need to find - \(a\) is the acceleration, which is the acceleration due to gravity (\(-9.81 \mathrm{~m/s^2}\)) - \(s\) is the displacement(vertical distance covered), which is given as 1/4 of the object's maximum height
03

Rewrite the displacement

The maximum height of an object thrown vertically is reached when its final velocity, \(v_f\), becomes 0. We can use the kinematic equation again: $$ v_f^2 = u^2 + 2as $$ to find the maximum height, \(s_{max}\), using the initial velocity \(u\) that we are looking for, the acceleration due to gravity, \(a = -9.81 \mathrm{~m/s^2}\), and displacement, \(s_{max}\). As \(v_f=0\), we have $$ 0 = u^2 - 2 \times 9.81 \times s_{max} $$ This means $$ s_{max} = \frac{u^2}{2 \times 9.81} $$ We are also given that the object has a velocity of \(25 \mathrm{~m/s}\) when it reaches 1/4 of its maximum height, \(s = \frac{1}{4}s_{max}\).
04

Write down equation for given velocity

Now we can write the equation for the given velocity \(v=25 \mathrm{~m/s}\): $$ (25 \mathrm{~m/s})^2 = u^2 - 2 \times 9.81 \times \frac{1}{4}s_{max} $$ Substitute the expression for \(s_{max}\) as found in Step 3: $$ (25 \mathrm{~m/s})^2 = u^2 - 2 \times 9.81 \times \frac{1}{4} \times \frac{u^2}{2 \times 9.81} $$
05

Solve for the initial velocity

Simplify the equation and solve for \(u\): $$ (25 \mathrm{~m/s})^2 = u^2 - \frac{1}{2}u^2 $$ $$ (25 \mathrm{~m/s})^2 = \frac{1}{2}u^2 $$ $$ u^2=\frac{2\times(25 \mathrm{~m/s})^2}{1} $$ Therefore, $$ u= \sqrt{\frac{2\times(25 \mathrm{~m/s})^2}{1}} = 35.36 \mathrm{~m/s} $$ So, the initial (launch) speed of the object is approximately \(35.36 \mathrm{~m/s}\).

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Key Concepts

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

Vertical Motion
Vertical motion refers to the movement of an object in a straight up-and-down path, typically under the influence of gravity. An important aspect is that the motion occurs only in the vertical direction, meaning there are no horizontal components to the trajectory. This means the equations used to describe the motion are focused solely on vertical displacement, velocity, and acceleration.

In the context of kinematics, vertical motion can be analyzed using the kinematic equations, which relate the parameters of velocity, acceleration, displacement, and time. This type of motion is often encountered when dealing with projectiles or any objects thrown vertically. When an object is tossed or launched upwards, it will rise until it reaches a point where its velocity is zero, also known as the maximum height. From there, the object will begin to descend, accelerating under gravity's pull.
Initial Velocity Calculation
Calculating the initial velocity of an object in vertical motion is crucial for understanding its motion trajectory. This calculation often uses the kinematic equations, which encompass a group of equations that describe motion in terms of initial velocity, final velocity, acceleration, time, and displacement.

To find the initial velocity in the given problem, we employ the equation:
  • \(v^2 = u^2 + 2as\)
In this equation:
  • \(v\) = final velocity at a certain point.
  • \(u\) = initial velocity, which is what we are trying to find.
  • \(a\) = acceleration due to gravity.
  • \(s\) = vertical displacement from the starting point.
By substituting the known values for final velocity (25 m/s), acceleration due to gravity, and displacement as a fraction of the maximum height, we can solve for the initial velocity. This approach will provide insight into how quickly the object was launched upward.
Acceleration due to Gravity
The acceleration due to gravity is a fundamental aspect when analyzing vertical motion. It is a constant that represents the rate at which velocity changes as a free-falling object moves under the influence of Earth's gravitational force.

On Earth, this constant is approximately \\(-9.81 \mathrm{~m/s^2}\). The negative sign indicates that the acceleration vector points downwards towards the Earth. This value is critical in vertical motion equations, determining how quickly an object's speed increases or decreases in the vertical direction.

In kinematic equations, the acceleration due to gravity is used to calculate various elements of motion, such as:
  • Maximum height reached by an object.
  • The time it takes for an object to reach that height.
  • The overall motion trajectory.
Understanding gravitational acceleration is essential for predicting how objects move when thrown or dropped, providing key insights into kinetic and potential energies in motion.

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Most popular questions from this chapter

The planet Mercury has a mass that is \(5 \%\) of that of Earth, and its gravitational acceleration is \(g_{\text {mercury }}=3.7 \mathrm{~m} / \mathrm{s}^{2}\) a) How long does it take for a rock that is dropped from a height of \(1.75 \mathrm{~m}\) to hit the ground on Mercury? b) How does this time compare to the time it takes the same rock to reach the ground on Earth, if dropped from the same height? c) From what height would you have to drop the rock on Earth so that the fall- time on both planets is the same?

Two cars are traveling at the same speed, and the drivers hit the brakes at the same time. The deceleration of one car is double that of the other. By what factor does the time required for that car to come to a stop compare with that for the other car?

A runner of mass \(56.1 \mathrm{~kg}\) starts from rest and accelerates with a constant acceleration of \(1.23 \mathrm{~m} / \mathrm{s}^{2}\) until she reaches a velocity of \(5.10 \mathrm{~m} / \mathrm{s}\). She then continues running at this constant velocity. How long does the runner take to travel \(173 \mathrm{~m} ?\)

Two train cars are on a straight, horizontal track. One car starts at rest and is put in motion with a constant acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\). This car moves toward a second car that is \(30.0 \mathrm{~m}\) away and moving at a constant speed of \(4.00 \mathrm{~m} / \mathrm{s}\). a) Where will the cars collide? b) How long will it take for the cars to collide?

Which of these statement(s) is (are) true? 1\. An object can have zero acceleration and be at rest. 2\. An object can have nonzero acceleration and be at rest. 3\. An object can have zero acceleration and be in motion. a) 1 only b) 1 and 3 c) 1 and 2 d) \(1,2,\) and 3
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