Projectile motion problem – determining initial velocity of throw

AI Thread Summary
The discussion revolves around solving a projectile motion problem involving a baseball thrown by a pitcher. The key approach involves using the kinematic equation s = ut + 0.5 at^2 to first determine the time of flight based on horizontal motion, followed by calculating the initial velocity. A participant points out inconsistencies in the use of signs for angles and distances, suggesting that the angle should be treated as positive to align with the positive downward direction for other quantities. They recommend simplifying the calculations by substituting time directly into the vertical motion equation to find the initial velocity more efficiently. The correct initial velocity of the pitch is confirmed as 45.2 m/s.
Kaiser98
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Homework Statement
A pitcher throws an overhand fastball from an approximate height of 2.65 m and at an
angle of 2.5° below horizontal. The catcher, catches the ball high in the strike zone, at a
height of 1.02 m above the ground. If pitcher's mound and home plate are 18.5 m apart,
what is the initial velocity of the pitch?
Relevant Equations
s = ut + 0.5 at^2

Answer given in book: 45.2 m/s
Screenshot 2022-11-28 at 4.11.00 AM.png


My reasoning was to use this kinematic equation to first get time of flight of the baseball using horizontal components, and then use this same equation again to find initial velocity.
 
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Kaiser98 said:
Homework Statement:: A pitcher throws an overhand fastball from an approximate height of 2.65 m and at an
angle of 2.5° below horizontal. The catcher, catches the ball high in the strike zone, at a
height of 1.02 m above the ground. If pitcher's mound and home plate are 18.5 m apart,
what is the initial velocity of the pitch?
Relevant Equations:: s = ut + 0.5 at^2

Answer given in book: 45.2 m/s

View attachment 317812

My reasoning was to use this kinematic equation to first get time of flight of the baseball using horizontal components, and then use this same equation again to find initial velocity.
Hello @Kaiser98 .
:welcome:

You are inconsistent with your use of signs.

You have vertical distance the ball drops as being positive as well as acceleration due to gravity as being positive. That's all fine.
However, you have the angle below horizontal as being negative. Since you had been treating the downward direction as being positive for the other quantities, so should the 2.5° angle be positive.
 
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Likes PeroK and Kaiser98
Also, your solution uses unnecessary algebra and the more you do of that, the more you expose yourself to mistakes. Why not substitute ##t=18.5/(u\cos(2.5^o))## in both occurrences of ##t## in the second (vertical) equation and solve directly for ##u##?
 
I did what both of you said, and finally got the right answer. Thanks a lot!
 
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