When is the vertical displacement negative in projectile motion equations?

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Homework Help Overview

The problem involves projectile motion, specifically analyzing the vertical displacement of a shell fired from a cliff. The shell's initial conditions include a height of 36m and a muzzle speed of 80.0m/s at an angle of 25 degrees above the horizontal. The original poster is attempting to demonstrate the time of flight using the vertical displacement equation.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster questions the sign of the vertical displacement in their equation, seeking clarity on when it should be negative. They express confusion over the placement of the displacement term in the equation.

Discussion Status

Participants are engaged in clarifying the correct application of signs in the equations related to vertical displacement, acceleration, and initial velocity. There is an exploration of how different coordinate systems can affect the signs of these terms, with some guidance provided on maintaining consistency among them.

Contextual Notes

Participants note that the choice of coordinate system is crucial, as it influences the signs of displacement, velocity, and acceleration. The original poster's confusion stems from the transition of the displacement term between sides of the equation and its implications based on the chosen reference frame.

totomyl
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Homework Statement


A shell is fired from a cliff that is 36m above a horizontal plane. the muzzle speed of the shell is 80.0m/s and it is fired at an elevation of 25 degrees above the horizontal. a) show that thetime of flight is 7.84 seconds.

Homework Equations


dy = viy(t) + 0.5(a)(t2)

The Attempt at a Solution


i got yp to
0 = 0.5(-9.8)(t2) + (80sin25)(t) - (36)

now, i know that i msut use the quadratic eq. but why is that -36m wrong in the equation, in my corrections it says it should be a positive 36 and be a -36 when it is on the left side of the equation. How do i know when to make the vertical displacement negative?
 
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-36 is correct, but it looks like you put it on the wrong side of the equation.

The displacement and the acceleration and velocities have to be in agreement with each other.
 
TomHart said:
-36 is correct, but it looks like you put it on the wrong side of the equation.
sorry for the confusion, i meant that i already moved the 36 from the left side to the right side making it -36, however that is apparently not correct, as it is -36 from the left side to begin with then becomes positive on the right side. why is that?
 
The terms have to be in agreement with each other. In other words, the initial velocity is in the opposite direction of the acceleration and displacement. So if you are calling acceleration negative, then the displacement has to be negative and the initial velocity has to be positive. Of course, you could reverse all of them if you wanted to.
 
It's up to you to choose your coordinate system before writing your equations. If you set the origin at the launch point with y increasing upwards then the final elevation (when the shell hits the ground below the cliff) will be negative. If you choose y increasing downwards then the change in elevation is positive, but you then need to change the signs of the velocity and acceleration to accommodate the choice of axes direction. If you choose the ground level for y = 0 then you need to incorporate the initial elevation (height of the cliff) into your equation.
 

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