Intro Physics projectile problem: Archer fish and grasshopper

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Homework Help Overview

The discussion revolves around a projectile motion problem involving an archer fish and a grasshopper, where participants are trying to determine the correct initial speed required to achieve a vertical displacement of 0.45 meters. There is a discrepancy between the calculated speed and the textbook answer.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculations related to vertical and horizontal components of velocity, questioning the assumption of initial velocity being zero and the relevance of the angle of projection in the calculations.

Discussion Status

Some participants have provided guidance on the importance of considering the angle of projection, which led to a recalculation resulting in a speed of 6 m/s. There is an ongoing exploration of the assumptions made in the initial calculations.

Contextual Notes

Participants note the absence of air resistance and the need for further information regarding the angle of projection, which appears to have been overlooked in earlier calculations.

Soccer
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HOMEWORK SUBMITTED IN WRONG FORUM, SO NO TEMPLATE

I uploaded an image of the problem, it seems to be fairly straight forward. However, my textbook indicates the answer as being 6 m/s, I am consistently getting 3 m/s. Any help would be appreciated!

Because there is no air resistance, the horizontal velocity is constant.

2ax+vo^2=vf^2
We can set the Vo equal to zero, find the required speed to achieve .45m vertically.
We find Viy=2.97m/s
vf=at
2.97=at
2.97/9.81=.3 seconds
Horizontal speed (distance over time)- constant speed
.25/.3=.833 m/s
(.833^2+2.97^2)^.5=3m/s

I know that the height and length indicated in the problem yield a angle of 61 degrees, however, this information still does not yield the correct answer.
 

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Soccer said:
HOMEWORK SUBMITTED IN WRONG FORUM, SO NO TEMPLATE

I uploaded an image of the problem, it seems to be fairly straight forward. However, my textbook indicates the answer as being 6 m/s, I am consistently getting 3 m/s. Any help would be appreciated!

Because there is no air resistance, the horizontal velocity is constant.

2ax+vo^2=vf^2
We can set the Vo equal to zero, find the required speed to achieve .45m vertically.
On what basis did you set vo equal to zero?
 
Chestermiller said:
On what basis did you set vo equal to zero?
You only need to calculate the speed to achieve 0.45m vertically, therefore it would be equivalent to the final speed of an object falling 0.45 meters vertically. Again, no air resistance.
 
Soccer said:
You only need to calculate the speed to achieve 0.45m vertically, therefore it would be equivalent to the final speed of an object falling 0.45 meters vertically. Again, no air resistance.
The velocity of the spit is not zero when it hits the bird. So neither the initial velocity nor the final velocity of the spit is zero.
 
There remains a constant x/horizontal component of velocity, the y vertical component of velocity can be zero at the point which the spit hits the object. Potentially, the y component of velocity could be nonzero at the top, however without any further information given, you can only assume the final y component of velocity is zero at the top.
 
Soccer said:
however without any further information given, you can only assume the final y component of velocity is zero at the top.
No.
You haven't used the given angle of projection in any of your calculations.
 
wow, I really appreciate that.. No idea why I decided to overlook the angle, got 6 m/s. Sorry for bothering you guys. Thank you!
 

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