## ParthKohli 2 years 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.

1. ParthKohli

look at this diagram

2. ParthKohli

|dw:1361876285295:dw|

3. ParthKohli

I dunno if that is right.

4. ParthKohli

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

5. RajshikharGupta

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

6. RajshikharGupta

7. RajshikharGupta

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

8. ParthKohli

It means exactly what it says.

9. ParthKohli

@mathslover

10. 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)

11. mathslover

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

12. ParthKohli

OK, we are assuming $$g = 9.8 ~m/s^2$$

13. ParthKohli

No, $$9.81$$

14. mathslover

ok

15. nincompoop

|dw:1361877650810:dw|

16. ParthKohli

$500 = \dfrac{u^2 }{9.81}$

17. ParthKohli

$u = \sqrt{500 \cdot 9.81}$

18. mathslover

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

19. mathslover

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

20. nincompoop

I think you two are making this way more complicated

21. Opcode

22. 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.

23. mathslover

I get it as 125 ft. (aprox.)

24. mathslover

and exactly it is 124.9

25. nincompoop

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

26. ParthKohli

38.069 meters

27. RajshikharGupta

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

28. mathslover

Yeah @rajshikhargupta you're also right . I think he might have not given view on that. Though , good to see you knew that :)

29. 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...

30. agent0smith

but, seriously, |dw:1367318591146:dw|

31. ParthKohli

lulz

32. nincompoop

hey now...

33. ruthjennifer

$R=\frac{ Vo ^{2}\sin 2\theta }{ g }$ $Vo ^{2}=\frac{ Rg }{ \sin 2\theta }$=$[152.4m(9.81s ^{2}/m)]/[\sin(2*45)]$=$1495.044 s ^{2}$ $H=\frac{Vo ^{2} \sin ^{2}θ }{ 2g }$ =$\frac{ {(1495.044 s ^{2})[\sin(45)^{2}]} }{2*9.81 \frac{ m }{ s ^{2} } }$ =38.1 m <------- answer