mechanics- Projectile motion (1 Viewer)

[vds700]

attached below.

I found all the equations of motion of the 2 stones, then the gradient of the line joining the 2stones in flight (to be tan<meta http-equiv="Content-Type" content="text/html; charset=utf-8"><meta name="ProgId" content="Word.Document"><meta name="Generator" content="Microsoft Word 11"><meta name="Originator" content="Microsoft Word 11"><link rel="File-List" href="file:///C:%5CUsers%5CAndrew%5CAppData%5CLocal%5CTemp%5Cmsohtml1%5C01%5Cclip_filelist.xml"><!--[if gte mso 9]><xml> <w:WordDocument> <w:View>Normal</w:View> <w:Zoom>0</w:Zoom> <wunctuationKerning/> <w:ValidateAgainstSchemas/> <w:SaveIfXMLInvalid>false</w:SaveIfXMLInvalid> <w:IgnoreMixedContent>false</w:IgnoreMixedContent> <w:AlwaysShowPlaceholderText>false</w:AlwaysShowPlaceholderText> <w:Compatibility> <w:BreakWrappedTables/> <w:SnapToGridInCell/> <w:WrapTextWithPunct/> <w:UseAsianBreakRules/> <wontGrowAutofit/> </w:Compatibility> <w:BrowserLevel>MicrosoftInternetExplorer4</w:BrowserLevel> </w:WordDocument> </xml><![endif]--><!--[if gte mso 9]><xml> <w:LatentStyles DefLockedState="false" LatentStyleCount="156"> </w:LatentStyles> </xml><![endif]--><style> <!-- /* Style Definitions */ p.MsoNormal, li.MsoNormal, div.MsoNormal {mso-style-parent:""; margin:0cm; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:"Times New Roman"; mso-fareast-font-family:"Times New Roman";} @page Section1 {size:595.3pt 841.9pt; margin:72.0pt 90.0pt 72.0pt 90.0pt; mso-header-margin:35.4pt; mso-footer-margin:35.4pt; mso-paper-source:0;} div.Section1 {page:Section1;} --> </style><!--[if gte mso 10]> <style> /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:10.0pt; font-family:"Times New Roman"; mso-ansi-language:#0400; mso-fareast-language:#0400; mso-bidi-language:#0400;} </style> <![endif]-->α ???) amd also the equarion of the line joining the 2 stones and subbing in values, but didn't work.

Can someone please help

 

[lolokay]

I can't really figure out what the question is asking. I get tanα for the gradient joining the lines too.
Is an "inclination to the horizontal distance from P to the foot of the wall" different from an inclination to the horizontal?

 

[vds700]

I can't really figure out what the question is asking. I get tanα for the gradient joining the lines too.
Is an "inclination to the horizontal distance from P to the foot of the wall" different from an inclination to the horizontal?

yeah i was a bit confused by that. I assumed that its the angle the line makes with the horizontal

 

[vds700]

There is a mistake in the question, i found the original paper and attached the REAL question below, Stupid CSSA or whoever typed these Q's up!

Lolokay and I were right with the first part though, tanα is independent of time.

And the next bit is easy, the distance is hcotα.

I'll have a go with b) and see how i go.


EDIT: Can someone show me how to do b). I get V(tanα + tanβ) =Vtanα lol

 

[lolokay]

for b) it's
y = v_yt - g/2 t²
dy/dt = v_y - gt

x = v_xt
dx/dt = v_x

so for a constant t
dy/dx = (v_y - gt)/v_x

so for 2 different t
(dy/dx)₁ - (dy/dx)₂ = -gt₁/v_x + gt₂/v_x

in the question the initial and final angles of the 2 projectiles are the same time apart, so
Ucosα(tanα - -tanα) = -g(t1 - t2)
Vcosα(tanα - -tanB) = -g(t1 - t2)
2Utanα = V(tanα + tanB)


edit: for the deduce part I keep getting U < V/4 :/

 

[vds700]

for b) it's
y = v_yt - g/2 t²
dy/dt = v_y - gt

x = v_xt
dx/dt = v_x

so for a constant t
dy/dx = (v_y - gt)/v_x

so for 2 different t
(dy/dx)₁ - (dy/dx)₂ = -gt₁/v_x + gt₂/v_x

in the question the initial and final angles of the 2 projectiles are the same time apart, so
Ucosα(tanα - -tanα) = -g(t1 - t2)
Vcosα(tanα - -tanB) = -g(t1 - t2)
2Utanα = V(tanα + tanB)


edit: for the deduce part I keep getting U < V/4 :/

could u please explain how u got the lines in bold- i don't follow what u say about the initial and final angles being the same time apart

Thanks for your help

 

[lolokay]

by "the same time apart" I just mean that t1 - t2 is the same for both of them, since the angles are taken at the same point in time

+ the Ucosa and Vcosa come from the v_x and the -tana and -tanB are negative since they're downwards angles

 

[vds700]

by "the same time apart" I just mean that t1 - t2 is the same for both of them, since the angles are taken at the same point in time

+ the Ucosa and Vcosa come from the v_x and the -tana and -tanB are negative since they're downwards angles

ah that makes more sense now. Thanks so much