anonymous
  • 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.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Going to try the problem until someone helps me.
anonymous
  • anonymous
|dw:1439370371073:dw|
anonymous
  • anonymous
is my sketch right?

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More answers

Jack1
  • Jack1
who cares if it's right... that sketch is epic art dude! props~!!!
anonymous
  • anonymous
especially the stickman? XD
Jack1
  • Jack1
totes ;)
Jack1
  • 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
  • anonymous
First thing I thought about this problem
anonymous
  • anonymous
component method. The vy= vosintheta
anonymous
  • anonymous
and vx=vocostheta
anonymous
  • anonymous
I assume I'll get the magnitude? Then I'll use the formula Vy = Voy - gt ?
anonymous
  • anonymous
Ohhh wait that's wrong
anonymous
  • anonymous
First thing is I need to know the y-component
anonymous
  • anonymous
|dw:1439371607809:dw|
Jack1
  • Jack1
|dw:1439371463873:dw|
anonymous
  • anonymous
Hmmm.. Do you I think I need to get the max height? (if it is possible)
Jack1
  • Jack1
i guess final height = 2.5m initial height = 0 m max height = ?
anonymous
  • anonymous
Jack1
  • Jack1
distance horizontal = 10m... initial speed = ?? acceleration vertical = -9.8 m/s i guess that's all we have to go on... yeah?
anonymous
  • anonymous
Yeaaap soooo
anonymous
  • anonymous
I posted a formula but I don't know if we can use it due to Vo is unknown too :(
Jack1
  • 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
  • anonymous
\[y=Vot-1/2(g)(t)^2\]
anonymous
  • anonymous
Okayy you have other ways I'm quite confuse
Jack1
  • Jack1
give me 15... just have to dinner, then we'll solve
anonymous
  • anonymous
Will do. :D
anonymous
  • anonymous
Another viewer welcome!!
Jack1
  • 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
  • anonymous
okay game
Jack1
  • 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
  • Jack1
what a freakin mouthful that is! ;D so... can you solve that in terms of Vo from here?
Jack1
  • Jack1
ooo... also: does that make sense? any issues with those kinematic equations dude?
anonymous
  • anonymous
Nah nah I'm good will read your text first gimme a sec
anonymous
  • anonymous
not literally a sec but maybe longer lol
Jack1
  • Jack1
|dw:1439376204600:dw|
anonymous
  • anonymous
wait still reading and understanding
anonymous
  • anonymous
Question about the distance equation
anonymous
  • anonymous
sy = Voy * t + 1/2 ay t^2 isn't it sy= Voy*t - 1/2g(t)^2 ??
anonymous
  • anonymous
or maybe u just translate it to other terms and it is just the same
anonymous
  • anonymous
nvm the question I got it
Jack1
  • Jack1
yep, its the same, i just use a for acceleration rather than g for gravity
anonymous
  • anonymous
in freefall ax is always 0 right?
anonymous
  • anonymous
bc there's force upon it only the y-component
anonymous
  • anonymous
no force*
Jack1
  • Jack1
yep, we assume wind resistance = negligable, so no x component to acceleration
anonymous
  • anonymous
Can you explain the logic behind putting t=10/voxcos38 in the sy=voyt+1/2ay(t)^2
Jack1
  • 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
  • anonymous
2.5 = (Vo x sin 38) * (10/(Vo * cos 38)) + 1/2 * -9.8 * (10/(Vo * cos 38))^2
anonymous
  • anonymous
isn't it that vo is also vox you just forgot to put x on it?
anonymous
  • anonymous
okay we're finding vox after this
anonymous
  • anonymous
no wait vox = 10 right?
Jack1
  • Jack1
no, we're finding Vo Vox is the x component (horizontal component) of Vo we just want Vo
anonymous
  • anonymous
lol jk
anonymous
  • anonymous
sx = 10
anonymous
  • anonymous
moving on I get it
Jack1
  • Jack1
2.5 = (Vo * sin 38) * (10/(Vo * cos 38)) + 1/2 * -9.8 * (10/(Vo * cos 38))^2
anonymous
  • anonymous
Okay moving on after I got Vo Ill use this?
anonymous
  • anonymous
anonymous
  • anonymous
2(vo)sin(38)/g
Jack1
  • 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
  • anonymous
Oh yeah.
anonymous
  • anonymous
I forgot about that
anonymous
  • anonymous
The question is quite odd I thought we need to find the T first before Vo. But ohwell what works.. works
Jack1
  • Jack1
focus dude, solve Vo
anonymous
  • anonymous
Ohmy I thought we are done. Sorry!! Forgive me. Wait a sec I'll solve
Jack1
  • Jack1
lol, sçool, jus wanna make sure we get the same answers for Vo and t, s'all ;)
anonymous
  • anonymous
I got vo=12.1871
Jack1
  • Jack1
perfect (+ or -... and minus is insane, so answer is positive 12.1871) now t?
anonymous
  • anonymous
1.0412 well that was fast
anonymous
  • anonymous
such a skilled swat team LOL
Jack1
  • Jack1
for a grenade... psh, not that fast like olympic sprinters = 100m in sub ten seconds... so yeah... sprint speed
anonymous
  • anonymous
well if you think of it under pressure and you're throwing it in a window! Such accuracy
Jack1
  • Jack1
totes well, i'm out, catchya dude, u cool with all that?
anonymous
  • anonymous
Of course thankyou!! U have yourself a new fan :D
Jack1
  • Jack1
logend, no worries bro, slaters ;P
Jack1
  • Jack1
*legend

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