ParthKohli
  • ParthKohli
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.
Physics
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ParthKohli
  • ParthKohli
look at this diagram
1 Attachment
ParthKohli
  • ParthKohli
|dw:1361876285295:dw|
ParthKohli
  • ParthKohli
I dunno if that is right.

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More answers

ParthKohli
  • ParthKohli
I don't know Physics. Can anyone help @abc9837?
anonymous
  • anonymous
or it would be h (max) the top point of projectile
anonymous
  • anonymous
what u all think about this!!..
anonymous
  • anonymous
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
ParthKohli
  • ParthKohli
It means exactly what it says.
ParthKohli
  • ParthKohli
@mathslover
mathslover
  • mathslover
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)
mathslover
  • mathslover
Oh sorry : R = \(\large{\frac{u^2 \sin 2\theta}{g}}\)
ParthKohli
  • ParthKohli
OK, we are assuming \(g = 9.8 ~m/s^2\)
ParthKohli
  • ParthKohli
No, \(9.81\)
mathslover
  • mathslover
ok
nincompoop
  • nincompoop
|dw:1361877650810:dw|
ParthKohli
  • ParthKohli
\[500 = \dfrac{u^2 }{9.81}\]
ParthKohli
  • ParthKohli
\[u = \sqrt{500 \cdot 9.81}\]
mathslover
  • mathslover
that will come approx. u = 50 root 2 or 70.7 [ i took g = 10 m/s^2 ]
mathslover
  • mathslover
and if we take g = 9.81 then it is u = 70.03 m/s
nincompoop
  • nincompoop
I think you two are making this way more complicated
Opcode
  • Opcode
Answer is zero.
mathslover
  • mathslover
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.
mathslover
  • mathslover
I get it as 125 ft. (aprox.)
mathslover
  • mathslover
and exactly it is 124.9
nincompoop
  • nincompoop
one unit is in meter and the other is in ft can you reconcile this?
ParthKohli
  • ParthKohli
38.069 meters
anonymous
  • anonymous
oh fishy what else i wrote in my first reply man!!! @parthkohli
mathslover
  • mathslover
Yeah @rajshikhargupta you're also right . I think he might have not given view on that. Though , good to see you knew that :)
agent0smith
  • agent0smith
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...
agent0smith
  • agent0smith
but, seriously, |dw:1367318591146:dw|
ParthKohli
  • ParthKohli
lulz
nincompoop
  • nincompoop
hey now...
anonymous
  • anonymous
\[R=\frac{ Vo ^{2}\sin 2\theta }{ g }\] \[Vo ^{2}=\frac{ Rg }{ \sin 2\theta }\]=\[[152.4m(9.81s ^{2}/m)]/[\sin(2*45)]\]=\[1495.044 s ^{2}\] \[H=\frac{Vo ^{2} \sin ^{2}θ }{ 2g }\] =\[\frac{ {(1495.044 s ^{2})[\sin(45)^{2}]} }{2*9.81 \frac{ m }{ s ^{2} } }\] =38.1 m <------- answer

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