Kinematics in Two Deminsions, Projectile Motion

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Homework Help Overview

The discussion revolves around a kinematics problem involving projectile motion, specifically related to an Olympic shotput event. The original poster is attempting to calculate the horizontal distance traveled by the shotput, given its initial velocity, launch angle, and height above the ground.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the setup of the equations for projectile motion, including the use of initial velocity components and gravitational effects. There are questions about the accuracy of calculations and the implications of rounding errors. Some participants suggest considering the sign of gravitational acceleration in the equations.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's approach and calculations. There is acknowledgment of potential errors and the importance of precision in calculations. Some participants express confidence in the setup while others question specific aspects of the calculations.

Contextual Notes

Participants note that the online system used for submitting answers may be sensitive to rounding errors, which could affect the final results. There is also mention of the need for careful consideration of signs in the equations used.

alexas
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Homework Statement


I have the angle and intial velocity but i need to find distance traveled in the x direction (i think).

In the Olympic shotput event, an athlete throws the shot with an initial speed of 12 m/s at a 40.0 (DEGREE) angle from the horizontal. The shot leaves her hand at a height of 1.8 m above the ground.

Known:
Angle (in degrees): 40
y inital: 1.8 m
Initial Velocity: 12 m/s
Intial Y compontent for velocity: 12sin(40) ?

Homework Equations


y= yo + voy – (0.5)g(t^2)


x = Vo(cos(40))t ?


The Attempt at a Solution



0 = 1.8m + 12(sin40)t + 1/2g(t^2)
Solve for zeros of t?
The only realistic one comes out to be about 1.7 since the other number is negative.
x = 12cos(40)t (plug t in) and get: 15.6m ?

The correct answer seems to be 16.6m so i know i must doing something wrong.
 
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You've set it up the way I would set it up, as far as I can tell. I don't have my calculator on me, so I'm not checking for silly math errors, but unless I'm missing something too, the set-up looks correct.

I've got an exam on this stuff Thursday, I hope I'm right in saying you set it up correctly.
 
alexas said:
0 = 1.8m + 12(sin40)t + 1/2g(t^2)
Solve for zeros of t?
The only realistic one comes out to be about 1.7 since the other number is negative.
x = 12cos(40)t (plug t in) and get: 15.6m ?

The correct answer seems to be 16.6m so i know i must doing something wrong.

I think you need to consider g is (-). But apparently you did to get 1.7 (Actually I get 1.782) If you had used +g you would have gotten 2 (-) times.

Perhaps if you carried greater precision through your calculation you would get a better answer?

Otherwise your method is fine.
 
LowlyPion said:
I think you need to consider g is (-). But apparently you did to get 1.7 (Actually I get 1.782) If you had used +g you would have gotten 2 (-) times.

Perhaps if you carried greater precision through your calculation you would get a better answer?

Otherwise your method is fine.

I basically just divide g by 1/2 and end up entering it (-4.9)

Also, when i take the 1.782 and put it into the second equation it doesn't equal 16.6?
 
Last edited:
Nevermind. The online system i am entering in the answers for is picking up on my round errors. Luck Me. =)
 
alexas said:
I basically just divide g by 1/2 and end up entering it (-4.9)

Also, when i take the 1.782 and put it into the second equation it doesn't equal 16.6?

Yes my point was that your equation showed a + sign instead of -g. As long as you keep it straight is what's important. Just be careful.

I also meant that you would use Sin and Cos to greater precision.

I see that you got it, and what's more important is that you understand how you got it, so ... all's well then.
 

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