A Projectile Fired At A 45 Degree Angle

  • Thread starter miniradman
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  • #26
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If you launch an identical projectile from 6 metres above the ground, with the same firing angle (45 deg), at 100m it will now clear a fence 12 metres high.
Touch'e :approve:

Ok, I'm lost... No matter what I do to the simultaneous equations. I never get the right answer :cry:

This is what I did

6=-9.81t^2 + y*(sin45 = 0.7071)----------------1
100=y*(cos45 = 0.7071)*t ----------------------2

make y the subject for eqn 2.

y=100/0.7071t

Sub y value into eqn 1

-6=-9.8t^2*(100/0.7071t)*0.7071

"0.7071" a "t" from -9.8t^t cancel out because of "0.7071" under the "100"

-6=-9.8t+100
100+6=9.8t
106/9.8 = 9.8t/9.8
9.24=t

I know I MUST have made a mistake but the worst part is that I cant see it :cry:
 
  • #27
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if the velocity of the y component is zero at 100 meters away and Vy=V0 - gt, then what does that tell you about the time it takes to get there?
hypothetically... if the y component was 0 at 100m, it will be at its peak height or on the ground. If it was at its peak height, this means that the time it takes to get there is affected by gravity. The only problem is that 100m mark isn't at peak height or on the ground.
 
  • #28
NascentOxygen
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-6=-9.8t^2*(100/0.7071t)*0.7071

Shouldn't the * be a + ?
 
  • #29
33
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yup, you switched a + for a *. also, it seems you added a minus to the 6. Drop the minus.

(I hope it wasn't because you misread my last post, the -6m thing was just an answer to your hypotethical situation where you stand 6m above ground and just clear a "fence 0 meters high".)
 
  • #30
195
0
I'll try it that way.

6=9.81t^2 + y*(sin45 = 0.7071)----------------1
100=y*(cos45 = 0.7071)*t ----------------------2

make y the subject for eqn 2.

y=100/0.7071t

Sub y value into eqn 1

6=9.8t^2*(100/0.7071t)*0.7071

"0.7071" a "t" from -9.8t^t cancel out because of "0.7071" under the "100"

6=9.8t+100
100-6=9.8t
94/9.8 = 9.8t/9.8

9.59=t

My answer should be around 4-5ish seconds
 
  • #31
33
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6=9.8t^2*(100/0.7071t)*0.7071

we just told you that you should have a plus in there, not a multiplication!

you do something wrong when you insert the substitution.
 
  • #32
195
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Ah, found the problem,

s = ut+(1/2) at^2

I used this formula and I got the right answer... thanks anyway guys
 
  • #33
33
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well you can't solve this problem with only one equation.
 
  • #34
195
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Yep, I used that s=ut+1/2 at^2 for both the horizontal and vertical components ;)
 
  • #35
33
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With 0 acceleration in the horionztal I take it. That's actually what I said in my equations.
 
  • #36
195
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Yes true, but there was still no 1/2 in either of them.

I tried to work it out without the 1/2 and I kept getting a cubic which cannot be solved. The one half made heaps of difference ;)
 
  • #37
33
0
OUCH! i simply forgot to write down the half! Pretty amazing that no-one noticed! Sorry about that, dude! :)


(the half actually appears from taking the anti-derivative of of t, resulting in 0.5(t)^2.)
 
Last edited:
  • #38
195
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I've haven't learnt about anti-derivatives yet. But I see what you have done there.

Its not a problem mate, everyone makes mistakes ;)
 
  • #39
33
0
Yup. Those 4 basic equations of motions is what you use when you start out with physics, but you see clearly where they naturally come from first when you learn about derivates and integration.

:)
 

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