Projectile Motion pea shooter question

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Homework Help Overview

The problem involves projectile motion, specifically analyzing the trajectory of a pea shot from a pea shooter in a classroom setting. The pea is fired a distance of 23 meters and hits a target 9.3 meters below the ceiling after 1.4 seconds, with the initial firing position noted as 3.1 meters below the ceiling.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial velocity in the vertical direction and the calculations related to vertical displacement. There are questions about the interpretation of the firing position and the relevance of horizontal distance in determining maximum height.

Discussion Status

Participants are actively engaging in clarifying the problem setup and addressing potential misunderstandings regarding the wording of the question. Some guidance has been offered regarding the separation of horizontal and vertical components of motion, but no consensus has been reached on the calculations or assumptions made.

Contextual Notes

There is mention of a worksheet provided by the teacher, which may contain unconventional wording that has led to confusion among participants. Additionally, there are references to potential sign errors in the calculations, indicating uncertainty in the approach taken.

Touran Khan
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Homework Statement


A student in the back row of class fires a pea shooter in an attempt to hit the blackboard. The student fires it a distance of 23m from his seat to hit the black board. The pea hits the board after 1.4 seconds at a location 9.3 meters below the ceiling. The pea is fired from a vertical position, 3.1m below the ceiling. Find the maximum height of the pea in its trajectory

Homework Equations



u = initial velocity
x = ut
y = ut + 1/2at2
v = u + at
v2 = u2 + 2ad

The Attempt at a Solution


Firstly, attempting to find initial velocity for y direction:

y direction:
d = 9.3 - 3.1 = 6.2 m
u = vsintheta
a = - 9.8 m/s2
t = 1.4 s

Using equations of motion: 6.2 = vsintheta + 1/2(-9.8)(1.4)2
vsintheta = 15.08 m/s

Finding max distance:

u = 15.08 m/s
v = 0 m/s
a = - 9.8 m/s

0 = 15.082 + 2(-9.8)(d)
d = 12.74 m

I feel like I've done something wrong? Not entirely sure, though. Don't have the answer to the questions. Thanks for the help!
 
Last edited:
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Touran Khan said:
The pea is fired from a vertical position, 3.1m below the ceiling.
Correction removed.
 
NascentOxygen said:
Horizontal position?

Yeah. My teacher gave us a worksheet full of questions that I presume he's made up, so I assumed that to be horizontal position as well.
 
Touran Khan said:
d = 9.3 - 3.1 = 6.2 m
Is that upwards or downwards?
 
Touran Khan said:
Yeah. My teacher gave us a worksheet full of questions that I presume he's made up, so I assumed that to be horizontal position as well.
As NascentO appears to have realized, it is only saying that the 3.1m refers to its vertical separation from the ceiling. The wording is a bit odd.
Touran Khan said:
vertical position, 3.1m below the ceiling
Is that comma there in the original or did you add it?
 
The pea is fired at an angle to the horizontal. You have used horiz distance and time to find ux.

Now use vert displacement and time to find uy.
 
haruspex said:
As NascentO appears to have realized, it is only saying that the 3.1m refers to its vertical separation from the ceiling. The wording is a bit odd.

Is that comma there in the original or did you add it?

Oh, alright. Honestly the wording of this question threw me off as well, but the comma was originally there. The question is exactly how my teacher has written it.
 
NascentOxygen said:
You have used horiz distance and time to find ux.
Seems to me Touran only calculated vertical motion, but made a sign error.
 
NascentOxygen said:
The pea is fired at an angle to the horizontal. You have used horiz distance and time to find ux.

Now use vert displacement and time to find uy.

How come we need the horizontal ux to find the max height of the projectile?
 
  • #10
Touran Khan said:
How come we need the horizontal ux to find the max height of the projectile?
You don't. NascentO seems to think that's what you did.
 
  • #11
haruspex said:
You don't. NascentO seems to think that's what you did.

Ah, apologies for the confusion. Also, which sign error have I made?
 
  • #12
Touran Khan said:
Ah, apologies for the confusion. Also, which sign error have I made?
Please answer my question in post #4.
 

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