How Do You Calculate Initial Velocity of an Arrow Shot Straight Up?

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Homework Help Overview

The discussion revolves around calculating the initial velocity of an arrow shot straight up, given its height after two seconds. The subject area is kinematics, specifically focusing on motion under the influence of gravity.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants are exploring the concept of initial velocity, questioning whether it refers to the velocity at launch or at maximum height. There is also discussion about the formula needed to find the initial velocity and the implications of the arrow's height after a specific time.

Discussion Status

Some participants have offered insights into the problem, with one suggesting a formula related to velocity and displacement. Others are questioning the completeness of the information provided and whether the initial velocity should be considered zero.

Contextual Notes

There is mention of a specific homework platform, which may impose constraints on how the problem is approached. Additionally, some participants express uncertainty about the sufficiency of the information given to solve the problem accurately.

swatmedic05
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You shoot an arrow into the air. Two seconds later (2.00 ) the arrow has gone straight upward to a height of 25.0 above its launch point
What is the initial velocity or the formula I would use to find
 
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Are you looking for max velocity? Initial velocity at what point? Before you shot it would be zero. At max height it would also be zero.
 
the right answer is 22.3 but i don't know how the book got that answer
 
can you post the entire question.
 
That is the entire question: My homework is on masteringphysics.com
 
There is not enough information to find the answer.
 
I apologize I think I just figured it out. Here is the formula
velocity & displacement:
v^2=initial velocity+2(-9.8m/s^2)(25m) but your initial velocity should be zero so are you sure that that is what the question is asking for.
 

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