Ball clearing fence

  • Thread starter apiwowar
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  • #1
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A batter hits a pitched ball when the center of the ball is 1.36 m above the ground.The ball leaves the bat at an angle of 45° with the ground. With that launch, the ball should have a horizontal range (returning to the launch level) of 112 m. (a) Does the ball clear a 7.65-m-high fence that is 102 m horizontally from the launch point? (b) At the fence, what is the distance between the fence top and the ball center?

so i solved for the initial velocity which is 33.13m/s
the i broke it up into its x and y components
Vx=23.43 m/s
Vy = 23.43 m/s

since the range is 112m i found the time that it takes for the ball to go 112m

t= x/Vx = 112m/23.43m/s

i then found the time it takes to 102 m using the same method and got 4.35s

so to find the height of the ball i did x=(23.43)(4.35)+1/2(9.8)(4.35)^2 = 9.2m

so i thought that it would be 2.91m above the fence but thats wrong.
where did i go wrong/?
 

Answers and Replies

  • #2
Doc Al
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Did you consider the initial height of the ball?
 
  • #3
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yes

to figure out the distance between the ball and the fence

i did (9.2+1.36)-7.65
 
  • #4
Doc Al
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Your work looks good to me. What book is this from?
 
  • #5
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halliday, fundamentals of physics 9e
its a wiley plus problem
 

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