Projectile motion of a baseball

  • Thread starter unknown_2
  • Start date
  • #1
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hey, i'm having some difficulty with this problem.
A baseball player makes perfect contact with the ball, striking it 45 degrees above the horizontal at a point 1.3m above the ground. his homerun hit just clears the 3.0m wall 130m from home plate. with what velocity did the baseball player strike the ball?
Code:
  /
 /                                                            |
[U]/ ) 45     [/U]                                                   |
    |                                                         | 3.0m 
    | 1.3m                                                    |
    |[U]                                                         |             [/U]    
                                 130m
this is pritty much what i have. i have no clue where to start. i've been trying to figure it out but no success... :(
 

Answers and Replies

  • #2
Shooting Star
Homework Helper
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Read up on projectile motion. Then ask any question.
 
  • #3
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uh...i did...i did lyk..30 questions on projectile motion...i know how to do the questions. i'm just stuck on this question.
 
Last edited by a moderator:
  • #4
12
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Were you not given a value for time between the ball being hit and it clearing the wall?

(Although you may not need it for this question)
 
  • #5
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ok...i am pretty sure i can help you witht his one :)
if you have been learning projectile motion in class then you should know:
[tex]\Delta Y = (V_{0}sin\varphi)(t)-.5gt^{2}[/tex]
where [tex]\varphi = 45 degrees (the angle of "launch"[/tex]
also you should know:
[tex]\Delta X = (V_{0}cos\varphi)(t)[/tex]

NOW here is what you will have to do:
solve the change in X function for t and plug that in for wherever "t" appears in the change in Y equation
set the change in Y to equal 1.7 m (the ball needs to at least hit that point to make it over the wall) and then solve that big nasty equation algebraically to get [tex]\ (V_{0} [/tex]

***edit : sorry its my first time with the math code so ill write this out
phi = 45 cuz its the angle of launch (the writing was kinda hard to read)
hope i was helpful!
 
Last edited:

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