Projectile motion of a baseball

AI Thread Summary
A baseball player hits a ball at a 45-degree angle from a height of 1.3 meters, needing it to clear a 3.0-meter wall located 130 meters away. The problem requires determining the initial velocity at which the ball was struck. To solve this, the equations for projectile motion are applied, specifically using the vertical and horizontal displacement equations. The vertical displacement must equal 1.7 meters (the difference between the wall height and the initial height) to ensure the ball clears the wall. By substituting the time variable from the horizontal equation into the vertical equation, the initial velocity can be calculated.
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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... :(
 
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Read up on projectile motion. Then ask any question.
 
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.
 
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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)
 
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:
\Delta Y = (V_{0}sin\varphi)(t)-.5gt^{2}
where \varphi = 45 degrees (the angle of "launch"
also you should know:
\Delta X = (V_{0}cos\varphi)(t)

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 \ (V_{0}

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