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An object is thrown vertically and has an upward velocity of 16 m/s when it reaches 1/2 of its maximum height above its launch point. What is the maximum height of the object?

An object is thrown vertically and has an upward velocity of 16 m/s when it reaches 1/2 of its maximum height above its launch point. What is the maximum height of the object?

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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**Problem Statement:** An object is thrown vertically and has an upward velocity of 16 m/s when it reaches 1/2 of its maximum height above its launch point. What is the maximum height of the object? --- To solve this problem, we can use the principles of kinematics and energy conservation. Here's a breakdown of the approach: 1. **Identify Known Variables:** - Initial velocity at half maximum height (v) = 16 m/s - Gravitational acceleration (g) = 9.8 m/s² 2. **Use the Kinematic Equation:** - \( v^2 = u^2 + 2as \) - Here, \( v \) is the velocity at half of the maximum height, \( u \) is the initial velocity, \( a \) is acceleration due to gravity (negative since it's acting downwards), and \( s \) is the distance traveled. 3. **Apply Energy Conservation:** - At maximum height, the velocity is 0. - The kinetic energy at half the height is equal to the potential energy gained. By using these principles, we can calculate the maximum height the object reaches.
Transcribed Image Text:**Problem Statement:** An object is thrown vertically and has an upward velocity of 16 m/s when it reaches 1/2 of its maximum height above its launch point. What is the maximum height of the object? --- To solve this problem, we can use the principles of kinematics and energy conservation. Here's a breakdown of the approach: 1. **Identify Known Variables:** - Initial velocity at half maximum height (v) = 16 m/s - Gravitational acceleration (g) = 9.8 m/s² 2. **Use the Kinematic Equation:** - \( v^2 = u^2 + 2as \) - Here, \( v \) is the velocity at half of the maximum height, \( u \) is the initial velocity, \( a \) is acceleration due to gravity (negative since it's acting downwards), and \( s \) is the distance traveled. 3. **Apply Energy Conservation:** - At maximum height, the velocity is 0. - The kinetic energy at half the height is equal to the potential energy gained. By using these principles, we can calculate the maximum height the object reaches.
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