What am I doing wrong in this basic Projectile Motion Problem?

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Homework Help Overview

The discussion revolves around a basic projectile motion problem, specifically focusing on the equations of motion for both horizontal and vertical displacements. Participants are examining the application of kinematic equations in the context of a stone thrown vertically upward.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations for horizontal and vertical motion, questioning the appropriateness of the equations applied by the original poster. There is a focus on the initial conditions, such as initial horizontal velocity and acceleration.

Discussion Status

Some participants have pointed out potential errors in the equations used by the original poster and have suggested corrections. There is an ongoing exploration of the implications of these corrections and the initial conditions of the problem, with no explicit consensus reached yet.

Contextual Notes

Participants are addressing the specific scenario of a stone thrown vertically upward, which raises questions about the horizontal displacement and the initial conditions of the motion. There is a mention of a typo in the equations presented, which may have contributed to confusion.

volleyballkay
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What am I doing wrong in this basic Projectile Motion Problem??

What I did in was use Delta X = Vxi + axt^2 & Delta Y= Vxi + ayt^2

then I plugged in the information but acceleration for x was 0m/s^2 and acceleration for y was -9.80m/s^2.

(ATTACHED AS FILE)
 

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Welcome to PF!

volleyballkay said:
What I did in was use Delta X = Vxi + axt^2 & Delta Y= Vxi + ayt^2

then I plugged in the information but acceleration for x was 0m/s^2 and acceleration for y was -9.80m/s^2.

(ATTACHED AS FILE)

I don't know whether this is the only thing you are doing wrong, but it does seem like there are a few errors with your equations. It should be:

Δx = vxit + (1/2)axt2

The parts you were missing are shown in bold.
 
Last edited:


volleyballkay said:
What I did in was use Delta X = Vxi + axt^2 & Delta Y= Vxi + ayt^2

then I plugged in the information but acceleration for x was 0m/s^2 and acceleration for y was -9.80m/s^2.

The stone was thrown vertically upward. If Δx means horizontal displacement, the initial horizontal velocity is zero, the horizontal acceleration is zero, is Δx different from zero?
You need to work only with ΔY=Vyit + ay/2* t^2. ay = -9.8 m/s^2, that is correct. What is the initial velocity? What is ΔY when the stone strikes the ground?

ehild
 
ehild said:
You meant Δx = vxit + (1/2)axt2, did you not? :smile:


ehild


Yeah I did. That was a very unfortunate typo.
 


Edited
 

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