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

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

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

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

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

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

RajshikharGupta
 one year ago
Best ResponseYou've already chosen the best response.0can 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
 one year ago
Best ResponseYou've already chosen the best response.1It means exactly what it says.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

ruthjennifer
 one year ago
Best 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
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