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ParthKohli
Group Title
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.
 one year ago
 one year ago
ParthKohli Group Title
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.
 one year ago
 one year ago

This Question is Closed

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
look at this diagram
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
dw:1361876285295:dw
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
I dunno if that is right.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
I don't know Physics. Can anyone help @abc9837?
 one year ago

RajshikharGupta Group TitleBest ResponseYou've already chosen the best response.0
or it would be h (max) the top point of projectile
 one year ago

RajshikharGupta Group TitleBest ResponseYou've already chosen the best response.0
what u all think about this!!..
 one year ago

RajshikharGupta Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
It means exactly what it says.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
@mathslover
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.3
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)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.3
Oh sorry : R = \(\large{\frac{u^2 \sin 2\theta}{g}}\)
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
OK, we are assuming \(g = 9.8 ~m/s^2\)
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
No, \(9.81\)
 one year ago

nincompoop Group TitleBest ResponseYou've already chosen the best response.1
dw:1361877650810:dw
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
\[500 = \dfrac{u^2 }{9.81}\]
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
\[u = \sqrt{500 \cdot 9.81}\]
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.3
that will come approx. u = 50 root 2 or 70.7 [ i took g = 10 m/s^2 ]
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.3
and if we take g = 9.81 then it is u = 70.03 m/s
 one year ago

nincompoop Group TitleBest ResponseYou've already chosen the best response.1
I think you two are making this way more complicated
 one year ago

Opcode Group TitleBest ResponseYou've already chosen the best response.0
Answer is zero.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.3
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.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.3
I get it as 125 ft. (aprox.)
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.3
and exactly it is 124.9
 one year ago

nincompoop Group TitleBest ResponseYou've already chosen the best response.1
one unit is in meter and the other is in ft can you reconcile this?
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
38.069 meters
 one year ago

RajshikharGupta Group TitleBest ResponseYou've already chosen the best response.0
oh fishy what else i wrote in my first reply man!!! @parthkohli
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.3
Yeah @rajshikhargupta you're also right . I think he might have not given view on that. Though , good to see you knew that :)
 one year ago

agent0smith Group TitleBest ResponseYou've already chosen the best response.0
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...
 one year ago

agent0smith Group TitleBest ResponseYou've already chosen the best response.0
but, seriously, dw:1367318591146:dw
 one year ago

nincompoop Group TitleBest ResponseYou've already chosen the best response.1
hey now...
 one year ago

ruthjennifer Group TitleBest ResponseYou've already chosen the best response.0
\[R=\frac{ Vo ^{2}\sin 2\theta }{ g }\] \[Vo ^{2}=\frac{ Rg }{ \sin 2\theta }\]=\[[152.4m(9.81s ^{2}/m)]/[\sin(2*45)]\]=\[1495.044 s ^{2}\] \[H=\frac{Vo ^{2} \sin ^{2}θ }{ 2g }\] =\[\frac{ {(1495.044 s ^{2})[\sin(45)^{2}]} }{2*9.81 \frac{ m }{ s ^{2} } }\] =38.1 m < answer
 one year ago
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