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ParthKohli

  • one year ago

Every once in a while a bird will get hit by a baseball during a baseball game. The longest home runs in baseball land about 500 feet (which is 152.4 meters) from home plate. If these balls leave the bat at a 45 degree angle to maximize their distance, what is the minimum safe height in meters for birds around baseball stadiums? On behalf of @abc9837, I have posted this problem.

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  1. ParthKohli
    • one year ago
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    look at this diagram

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  2. ParthKohli
    • one year ago
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    |dw:1361876285295:dw|

  3. ParthKohli
    • one year ago
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    I dunno if that is right.

  4. ParthKohli
    • one year ago
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    I don't know Physics. Can anyone help @abc9837?

  5. RajshikharGupta
    • one year ago
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    or it would be h (max) the top point of projectile

  6. RajshikharGupta
    • one year ago
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    what u all think about this!!..

  7. RajshikharGupta
    • one year ago
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    can u xplain what the hell this "Every once in a while a bird will get hit by a baseball during a baseball game" mean

  8. ParthKohli
    • one year ago
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    It means exactly what it says.

  9. ParthKohli
    • one year ago
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    @mathslover

  10. mathslover
    • one year ago
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    actually this question does not seemt oo much hard to me. What we are given with is horizontal range (R) and theta (angle) given : R = 500 ft. and theta = 45 degrees Since \(R = \large{\frac{u^2 \sin 2\theta}{2g}}\) = 500ft. put theta = 45 degrees and g = 10 m/s^2 . calculate the *u* first (initial velocity)

  11. mathslover
    • one year ago
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    Oh sorry : R = \(\large{\frac{u^2 \sin 2\theta}{g}}\)

  12. ParthKohli
    • one year ago
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    OK, we are assuming \(g = 9.8 ~m/s^2\)

  13. ParthKohli
    • one year ago
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    No, \(9.81\)

  14. mathslover
    • one year ago
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    ok

  15. nincompoop
    • one year ago
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    |dw:1361877650810:dw|

  16. ParthKohli
    • one year ago
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    \[500 = \dfrac{u^2 }{9.81}\]

  17. ParthKohli
    • one year ago
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    \[u = \sqrt{500 \cdot 9.81}\]

  18. mathslover
    • one year ago
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    that will come approx. u = 50 root 2 or 70.7 [ i took g = 10 m/s^2 ]

  19. mathslover
    • one year ago
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    and if we take g = 9.81 then it is u = 70.03 m/s

  20. nincompoop
    • one year ago
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    I think you two are making this way more complicated

  21. Opcode
    • one year ago
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    Answer is zero.

  22. mathslover
    • one year ago
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    now we have u as initial velocity calculate the max. height of the projection (H): \[\large{ H = \frac{ u^2 \sin ^2 \theta}{2g}}\] Calculate H now.

  23. mathslover
    • one year ago
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    I get it as 125 ft. (aprox.)

  24. mathslover
    • one year ago
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    and exactly it is 124.9

  25. nincompoop
    • one year ago
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    one unit is in meter and the other is in ft can you reconcile this?

  26. ParthKohli
    • one year ago
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    38.069 meters

  27. RajshikharGupta
    • one year ago
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    oh fishy what else i wrote in my first reply man!!! @parthkohli

  28. mathslover
    • one year ago
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    Yeah @rajshikhargupta you're also right . I think he might have not given view on that. Though , good to see you knew that :)

  29. agent0smith
    • one year ago
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    I don't remember how I got here, but that is an awesome drawing by @nincompoop and it deserved more attention and medals. Also, even though it's 2 months old, looks like you guys messed up your calculation @mathslover and @ParthKohli "given : R = 500 ft. put theta = 45 degrees and g = 10 m/s^2 " You both used the range in feet, but gravity in metres...

  30. agent0smith
    • one year ago
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    but, seriously, |dw:1367318591146:dw|

  31. ParthKohli
    • one year ago
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    lulz

  32. nincompoop
    • one year ago
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    hey now...

  33. ruthjennifer
    • one year ago
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    \[R=\frac{ Vo ^{2}\sin 2\theta }{ g }\] \[Vo ^{2}=\frac{ Rg }{ \sin 2\theta }\]=\[[152.4m(9.81s ^{2}/m)]/[\sin(2*45)]\]=\[1495.044 s ^{2}\] \[H=\frac{Vo ^{2} \sin ^{2}θ }{ 2g }\] =\[\frac{ {(1495.044 s ^{2})[\sin(45)^{2}]} }{2*9.81 \frac{ m }{ s ^{2} } }\] =38.1 m <------- answer

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