Integration and initial velocity

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Homework Help Overview

The problem involves determining the initial velocity required for an object to reach a maximum height of 550 feet when thrown upward, using the acceleration due to gravity as -32 ft/sec² and neglecting air resistance. The context is rooted in kinematics and integration.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration of acceleration to find velocity, with one noting confusion regarding the variable used in the integration. Questions arise about the meaning of the variable 'x' and its relationship to time, as well as the interpretation of the integration constant.

Discussion Status

The discussion is ongoing, with participants exploring the integration process and clarifying notation. Some guidance has been offered regarding the integration of acceleration and the implications for determining the initial conditions, but no consensus has been reached on how to proceed further.

Contextual Notes

There is a noted lack of precision in notation, particularly regarding the variable representing time, which may affect the clarity of the discussion. Participants are also considering how to determine the integration constant based on the conditions of the problem.

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Homework Statement



With what initial velocity must an object be thrown upward (from ground level) to reach a maximum height of 550feet.

Use a(t)= -32ft/sec2 as the acceleration due to gravity. (neglect air resistance)

Homework Equations



Use integration

The Attempt at a Solution



I know that U should first start off by integrating the acceleration in order to get velocity, but I wind up getting:

32x+C= v(t)

I'm not sure how to deal wit the problem from here, does anyone have any suggestions?
 
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What is x?
 
Im pretty sure that x is suppose to be time, because that is the only thing that relates acceleration and velocity
 
I thought t was time as well. So lesson one is to be a bit more precise in your notation.

a(t) = -32 ft/s2, then
v(t) = \int a(t) \, dt
where t is the variable. Integrating a constant over t gives you the constant times t so
v(t)[ft/s] = - 32 t + C

Note the minus sign, which is carried over from a(t) < 0.
C is an integration constant which you need to determine. What condition will you use for this?
How can you see in the v(t) graph or formula that the highest point is reached?
 

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