- anonymous

A SWAT team who is 10 meters away from the bus launches a tear gas that hits its wrecked window at point 2.5 meters above the ground. IF the tear gas is projected at an angle 38 degrees from the horizontal, determine its total time of flight. What is the initial velocity of the tear gas? Sketch the problem.

- schrodinger

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- anonymous

Going to try the problem until someone helps me.

- anonymous

|dw:1439370371073:dw|

- anonymous

is my sketch right?

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## More answers

- Jack1

who cares if it's right... that sketch is epic art dude!
props~!!!

- anonymous

especially the stickman? XD

- Jack1

totes ;)

- Jack1

10m away = yep
2.5 high window = yep
i guess we're assuming they launch the gas thing from ground level??
38 degrees from horizontal... yep
drawing looks great
now, the important thing: can u solve it?

- anonymous

First thing I thought about this problem

- anonymous

component method. The vy= vosintheta

- anonymous

and vx=vocostheta

- anonymous

I assume I'll get the magnitude? Then I'll use the formula Vy = Voy - gt ?

- anonymous

Ohhh wait that's wrong

- anonymous

First thing is I need to know the y-component

- anonymous

|dw:1439371607809:dw|

- Jack1

|dw:1439371463873:dw|

- anonymous

Hmmm.. Do you I think I need to get the max height? (if it is possible)

- Jack1

i guess
final height = 2.5m
initial height = 0 m
max height = ?

- anonymous

##### 1 Attachment

- Jack1

distance horizontal = 10m...
initial speed = ??
acceleration vertical = -9.8 m/s
i guess that's all we have to go on... yeah?

- anonymous

Yeaaap soooo

- anonymous

I posted a formula but I don't know if we can use it due to Vo is unknown too :(

- Jack1

so ur going to have to rephrase the time in terms of the horizontal distance, then plug that into ur max height calc... sounds like a mess of work...

- anonymous

\[y=Vot-1/2(g)(t)^2\]

- anonymous

Okayy you have other ways I'm quite confuse

- Jack1

give me 15... just have to dinner, then we'll solve

- anonymous

Will do. :D

- anonymous

Another viewer welcome!!

- Jack1

hey dude, back
so:
known factors:
initial height = 0
final height = 2.5m
max height = ?
initial distance = 0
final distance = 10m
acceleration:
ay = -9.8 m/s^2
ax = 0
Initial Speeds:
Voy = Vo x sin 38
Vox = Vo x cos 38
Distance equations:
sy = Voy * t + 1/2 ay t^2
sx = Vox * t + 1/2 ax t^2

- anonymous

okay game

- Jack1

sx = Vox * t + 1/2 ax t^2
and ax = 0
sx = Vox * t + 0
10 = Vox * t
t = 10/Vox
and
Vox = Vo * cos 38
so
t = 10/Vox
t = 10/(Vo * cos 38)
sy = Voy * t + 1/2 ay t^2
2.5 = Voy * t + 1/2 ay t^2
2.5 = Voy * (10/(Vo * cos 38)) + 1/2 * -9.8 * (10/(Vo * cos 38))^2
and
Voy = Vo x sin 38
so
2.5 = Voy * (10/(Vo * cos 38)) + 1/2 * -9.8 * (10/(Vo * cos 38))^2
2.5 = (Vo x sin 38) * (10/(Vo * cos 38)) + 1/2 * -9.8 * (10/(Vo * cos 38))^2

- Jack1

what a freakin mouthful that is! ;D
so... can you solve that in terms of Vo from here?

- Jack1

ooo... also: does that make sense? any issues with those kinematic equations dude?

- anonymous

Nah nah I'm good will read your text first gimme a sec

- anonymous

not literally a sec but maybe longer lol

- Jack1

|dw:1439376204600:dw|

- anonymous

wait still reading and understanding

- anonymous

Question about the distance equation

- anonymous

sy = Voy * t + 1/2 ay t^2
isn't it sy= Voy*t - 1/2g(t)^2 ??

- anonymous

or maybe u just translate it to other terms and it is just the same

- anonymous

nvm the question I got it

- Jack1

yep, its the same, i just use a for acceleration rather than g for gravity

- anonymous

in freefall ax is always 0 right?

- anonymous

bc there's force upon it only the y-component

- anonymous

no force*

- Jack1

yep, we assume wind resistance = negligable, so no x component to acceleration

- anonymous

Can you explain the logic behind putting t=10/voxcos38 in the sy=voyt+1/2ay(t)^2

- Jack1

because we want an equation with only 1 unknown in it... so we put everything in terms of inital velocity (Vo - not component split)
then we solve for Vo
then we use that to solve for time

- anonymous

2.5 = (Vo x sin 38) * (10/(Vo * cos 38)) + 1/2 * -9.8 * (10/(Vo * cos 38))^2

- anonymous

isn't it that vo is also vox you just forgot to put x on it?

- anonymous

okay we're finding vox after this

- anonymous

no wait vox = 10 right?

- Jack1

no, we're finding Vo
Vox is the x component (horizontal component) of Vo
we just want Vo

- anonymous

lol jk

- anonymous

sx = 10

- anonymous

moving on I get it

- Jack1

2.5 = (Vo * sin 38) * (10/(Vo * cos 38)) + 1/2 * -9.8 * (10/(Vo * cos 38))^2

- anonymous

Okay moving on after I got Vo Ill use this?

- anonymous

##### 1 Attachment

- anonymous

2(vo)sin(38)/g

- Jack1

just solve Vo first
then put that back into
t = 10/(Vo * cos 38)
to solve for t
then... well ... done, yeah?
ur time in flight equation assumes the start and finish height are the same... so it wont work for this question

- anonymous

Oh yeah.

- anonymous

I forgot about that

- anonymous

The question is quite odd I thought we need to find the T first before Vo. But ohwell what works.. works

- Jack1

focus dude, solve Vo

- anonymous

Ohmy I thought we are done. Sorry!! Forgive me. Wait a sec I'll solve

- Jack1

lol, sÃ§ool, jus wanna make sure we get the same answers for Vo and t, s'all ;)

- anonymous

I got vo=12.1871

- Jack1

perfect (+ or -... and minus is insane, so answer is positive 12.1871)
now t?

- anonymous

1.0412 well that was fast

- anonymous

such a skilled swat team LOL

- Jack1

for a grenade... psh, not that fast
like olympic sprinters = 100m in sub ten seconds... so yeah... sprint speed

- anonymous

well if you think of it under pressure and you're throwing it in a window! Such accuracy

- Jack1

totes
well, i'm out, catchya dude, u cool with all that?

- anonymous

Of course thankyou!! U have yourself a new fan :D

- Jack1

logend, no worries bro, slaters ;P

- Jack1

*legend

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