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Projectiles launched at an angle

  • Thread starter aly1201
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  • #1
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Homework Statement



A baseball is hit over a 7.5-m-high fence 95 m from the home plate. What was the minimum speed of the ball when it left the bat? Assume the ball was hit 1 m above the ground and its path initially made a 38 degree angle with the ground.

Homework Equations



y=v(initial, y direction)t-1/2at^2 x=v(initial, x direction)t

The Attempt at a Solution



t=95/v 95/v(cos38) ? plug 95/v(cos38) into the y equation? and substitute v(sin38) for initial y velocity? This is where I get lost; our teacher wrote all the work on the board very quickly and came up with an answer somehow, but when I did the work I didnt even come close.
 

Answers and Replies

  • #2
G01
Homework Helper
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You have the following (correct) equation for the height of the ball right when it gets to the fence:

[tex]y=v\sin(38^o)-\frac{1}{2}g\left(\frac{95}{v\cos(38^o)}\right)^2[/tex]

Now how high do you want the ball to be so that it just barely the fence? Plug that in for y and when you solve for v, you'll get the minimum speed.
 
  • #3
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Ok but I dont understand how to get v by itself, with the cos38 and sin38 there. And in the equation you wrote is "t" supposed to be in there at all?
 
  • #4
hage567
Homework Helper
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There is a t missing in the vsin(38) term in that equation. G01 has substituted for t since you can express it in terms of the horizontal distance and horizontal velocity. As already pointed out, you can figure out what y should be from the information given in the problem. So the equation should look like this:

[tex]y=v\sin(38^o)\left(\frac{95}{vcos(38^o)}\right)-\frac{1}{2}g\left(\frac{95}{v\cos(38^o)}\right)^2[/tex]

Now it is just algebra. To isolate v, start by getting the term containing v on its own on one side of the equation (note the v's cancel in the first term!)
 
  • #5
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Ok so [95/v(cos38)]^2=[6.5-(tan38)(95)]/-4.9? because sinX/cosX=tanX
 
  • #6
hage567
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Looks good so far. What would you do next?
 
  • #7
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I solved the right side and got 13.8, which i took the square root of (since the left side was squared) and got 3.7. Then I divided that by 95 and got 25.6. So now v(cos38)=25.6? If it does then I could just say 25.6/cos38=v, so 25.6/.79=32.4?
I might've rounded too much so it might not be exact, and Im not sure if you can have cosX or sinX by itself?
 
  • #8
hage567
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That answer looks OK to me.
 
  • #9
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It was thank you! I appreciate the help!
 

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