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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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look at this diagram
1 Attachment
|dw:1361876285295:dw|
I dunno if that is right.

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Other answers:

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

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