Finding muzzle velocity knowing the angle, height shot from & horizontal displacement


by Jvells
Tags: angle, displacement, height, horizontal, knowing, muzzle, shot, velocity
Jvells
Jvells is offline
#1
Dec1-12, 12:29 PM
P: 6
I need help with my hw, bad. I've been screwing off in my physics class for awhile and I can't get afterschool help because I have sports. I need to pick my grade up in this class to keep playing...so here I am. I need a good walkthrough badly, i don't know a whole lot with physics, but a decent amount and I'm begginggg someone to help guide me through this at least a little bit...

I need to figure out the initial velocity (muzzle velocity) of a ball after it got shot knowing the angle, height, and distance traveled.

He gave me these two formulas: "Δx=Vxt" and "Δy=Vyit+1/2ayt^2".

The projectile went: 0.98m
Launched horizontally: (0=0°). Why "0=" and not just 0°? Guess the 0 refers to the x axis?
Launched from the height: 0.79m

PLEASE HELP getting me started with this. It'll be muchhh appreciated...I have quite a bit to catch up on and this is the assignment I'm starting with.
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K^2
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#2
Dec1-12, 01:05 PM
Sci Advisor
P: 2,470
Could it have been "θ=0°"? That would make more sense. θ is a common symbol for angle measure. Not much to do with the problem, though. So anyways... Lets see if these hints help you.

1) Δx and Δy in these formulas refer to distances traveled between time 0 and t. So what are the Δx and Δy for this problem?

2) The projectile is launched horizontally. That tells you something about initial velocity. What variable can you replace or get rid of in your formulas based on this information?

Once you have these things, just plug everything you know into formulas and see if any of them can be solved for any of the unknowns.

By the way, normally, on this forum, you are expected to show as much of the work as you've managed on the problem before asking for help. Since you are saying that you are completely stuck, the above will hopefully act as the guided way to do the same.
Jvells
Jvells is offline
#3
Dec1-12, 02:20 PM
P: 6
yeah it was theta not 0, I looked at it wrong my bad. But yeah you did help a bit.

1) Δx would be 0.98m & Δy would be 0.79m?

2) Beign launched horizontally the initial velocity would be 0 since it's at rest before being shot. I'm really confused though so I could be wrong. Now how do I figure out the acceleration (ay)...? Would you mind doing a little math for me on here so I can see what you end up doing and figure it out that way...?

K^2
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#4
Dec1-12, 02:48 PM
Sci Advisor
P: 2,470

Finding muzzle velocity knowing the angle, height shot from & horizontal displacement


If there are no other forces acting on projectile, vertical acceleration is just acceleration due to gravity, which is about -9.8m/sē. So long as the problem is set up on Earth, this will always be the same.

You should try to keep your signs consistent. Since projectile started at height of 0.79m and ended up at height of 0m, Δy = -0.79m. (Final minus initial.)
Jvells
Jvells is offline
#5
Dec1-12, 07:22 PM
P: 6
I understand that much, now how do i go about solving this for t? Would you mind doing it for me and showing the work on here...? I'd loveee you. I'm not just trying to just get it over with, im gunna review your work and figure it out. That's if you do it, no pressuree.

So, here's what i need worked out: "-0.79m=(0)t+1/2(-9.8m/s^2)t^2". If you do it, thanks a ton man...
Jvells
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#6
Dec2-12, 07:55 AM
P: 6
Bump...
K^2
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#7
Dec2-12, 10:03 AM
Sci Advisor
P: 2,470
Quote Quote by Jvells View Post
So, here's what i need worked out: "-0.79m=(0)t+1/2(-9.8m/s^2)t^2". If you do it, thanks a ton man...
You can perform the same operation both sides of an equal sign, and equality remains valid. First of all, you can just get rid of 0*t, because that's just 0, and 0 + something is just that something. Next, you can multiply by (-1) on both sides, which will get rid of the minus sign. Finally, you want to take a square root of both sides. Keep in mind that [itex]\sqrt{Number*t^2} = \sqrt{Number}*t[/itex]. That should leave you with a simple linear equation which, I hope, you know how to solve for t.

By the way, for future, you might want to learn how to solve quadratic equations. Only so much of physics is physics. The rest is algebra, and quadratic equations will come back. So will simple systems of linear equations.
Jvells
Jvells is offline
#8
Dec2-12, 12:02 PM
P: 6
T=0.40 seconds

Rounded to two sig digs.

Solved the other equation for a final muzzle velocity of 2.5 m/s. Looks right to me, what do you say mann? Absolutely love you for helping, i get it.
Jvells
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#9
Dec3-12, 01:33 PM
P: 6
Bump...sorry
K^2
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#10
Dec3-12, 02:42 PM
Sci Advisor
P: 2,470
Yeah, these look good.


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