Projectile motion question (baseball problem)

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Homework Help Overview

The problem involves projectile motion, specifically analyzing the trajectory of a baseball hit at an angle. The baseball's initial height, launch angle, distance to a fence, and the height of the fence are provided, with the goal of determining the initial speed required for the baseball to clear the fence.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the components of the initial velocity, questioning the use of vertical and horizontal components in their calculations. There is consideration of whether to include the diameter of the baseball in the height calculation when determining if it clears the fence.

Discussion Status

Multiple interpretations of the problem are being explored, particularly regarding the correct application of projectile motion equations. Some participants have provided guidance on checking calculations and ensuring the correct use of trigonometric functions. There is no explicit consensus on the correct initial speed yet, as participants continue to refine their approaches.

Contextual Notes

Participants note the importance of precision in calculations, particularly regarding the height at which the baseball clears the fence and the potential impact of the ball's diameter on this measurement. There is also mention of homework constraints that may affect the problem-solving process.

boomer77
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Homework Statement



The center of a baseball of diameter 8.20 cm is 1.320 m vertically above the plate when it is hit. The blast sends it off at an angle of 30.5° above the horizontal with an unknown initial speed. The outfield fence is 94.0 m away and 11.20 m tall: the ball just clears it. Ignoring aerodynamic effects, what was the initial speed of the baseball?

Homework Equations



-d=vt for the x direction
-any free fall equation for the y direction

The Attempt at a Solution



dx= 94.0m dy= 9.88m (subtracted the height in which the ball was hit at)
Vx= vcos(30.5 Viy=?
t= 94.0/vcos(30.5 a= -9.8m.s^2

d=vit+1/2at^2
9.88=vi(94.0/vcos(30.5)-4.9(94.0/vcos(30.5)^2
v^2(-99.22)= -58323.769
v^2= 5884.13
v=76.7m/s

where am i going wrong in this problem? help will be greatly appreciated, thanks!
 
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You have Vx=vcostheta; how about Viy=vsintheta? Looks like you have missed that.
 
thanks for the reply. do i add the diameter of the ball to when it clears the fence too? I'm pretty bad with word problems.
 
boomer77 said:
thanks for the reply. do i add the diameter of the ball to when it clears the fence too? I'm pretty bad with word problems.
i guess to be accurate to 2 places after the decimal point, you'd have to consider the ball diameter, that is, add half its diameter to the y height (y = 9.88 + 0.04 = 9.92) for what it may be worth.
 
i'm still getting this problem wrong so my equation in the y direction would be

9.92=vsin(30.5(94/vcos(30.5)-4.9(94/vcos(30.5)^2

i come up with the answer of 3.88 m/s but my problem set says it's wrong

should i be doing something different?
 
boomer77 said:
i'm still getting this problem wrong so my equation in the y direction would be

9.92=vsin(30.5(94/vcos(30.5)-4.9(94/vcos(30.5)^2

i come up with the answer of 3.88 m/s but my problem set says it's wrong

should i be doing something different?
Looks like the correct approach; I'll check your numbers in a bit...I've got to pick up a take out order for the family..
 
boomer77 said:
i'm still getting this problem wrong so my equation in the y direction would be

9.92=vsin(30.5)[/color](94/vcos(30.5))[/color] - 4.9(94/vcos(30.5))[/color]^2

i come up with the answer of 3.88 m/s but my problem set says it's wrong

should i be doing something different?
Looks like a math error..I get V = 35.99m/s. Check your parentheses..check calculator set to degrees and not radians..
 
thank you very much for the help!
 

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