Projectile Motion pea shooter question

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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!
 
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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:
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?
 
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:
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?
 
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?