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.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- ParthKohli

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- ParthKohli

look at this diagram

##### 1 Attachment

- ParthKohli

|dw:1361876285295:dw|

- ParthKohli

I dunno if that is right.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- ParthKohli

I don't know Physics. Can anyone help @abc9837?

- anonymous

or it would be h (max) the top point of projectile

- anonymous

what u all think about this!!..

- 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

It means exactly what it says.

- ParthKohli

- 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

Oh sorry : R = \(\large{\frac{u^2 \sin 2\theta}{g}}\)

- ParthKohli

OK, we are assuming \(g = 9.8 ~m/s^2\)

- ParthKohli

No, \(9.81\)

- mathslover

ok

- nincompoop

|dw:1361877650810:dw|

- ParthKohli

\[500 = \dfrac{u^2 }{9.81}\]

- ParthKohli

\[u = \sqrt{500 \cdot 9.81}\]

- mathslover

that will come approx. u = 50 root 2 or 70.7 [ i took g = 10 m/s^2 ]

- mathslover

and if we take g = 9.81 then it is u = 70.03 m/s

- nincompoop

I think you two are making this way more complicated

- Opcode

Answer is zero.

- 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

I get it as 125 ft. (aprox.)

- mathslover

and exactly it is 124.9

- nincompoop

one unit is in meter and the other is in ft
can you reconcile this?

- ParthKohli

38.069 meters

- anonymous

oh fishy
what else i wrote in my first reply man!!!
@parthkohli

- 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

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

but, seriously,
|dw:1367318591146:dw|

- ParthKohli

lulz

- nincompoop

hey now...

- 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.