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Projectile Motion: Range proof 
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#1
Nov105, 03:02 PM

P: 110

Q. A projectile is fired with an initial speed [tex]V_0[/tex] at an angle [tex]\beta[/tex] to the horitontal. Show that it's range alnog a plane which it's self is inclined at an angle [tex]\alpha[/tex] to the horitontal [tex]( \beta > \alpha)[/tex] is given by:
[tex] R = \frac{(2{V_0}^2 cos \beta sin(\beta  \alpha )}{g {cos}^2\alpha} [/tex] A.So I`ve started off with [tex]\triangle x = (V_0 cos \beta) t[/tex] and [tex]\triangle y = (V_0 sin \beta) t  frac{1}{2} g t^2[/tex] [tex]\triangle x = cos \alpha[/tex]and [tex]\triangle y = sin \alpha[/tex] so i rearranged [tex]\triangle x[/tex] to get [tex]t = \frac{cos \alpha}{V_0 cos \beta}[/tex] and sub it into [tex]\triangle y[/tex] now i have [tex] \triangle y = (V_0 sin \beta)\frac{cos \alpha}{V_0 cos \beta}  \frac{1}{2} g ( \frac{cos \alpha}{V_0 cos \beta} )[/tex] [tex] \triangle y= \frac {V_0 sin \beta cos \alpha}{V_0 cos \beta}  \frac{1}{2} g ( \frac{cos \alpha}{V_0 cos \beta})[/tex] [tex] sin \alpha = tan \beta cos \alpha  \frac {g {cos}^2 \alpha}{2 {V_0}^2 {cos}^2 \beta} [/tex] now i`m stuck ... Any hint`s/tips would be great. Cheers. 


#2
Nov105, 03:24 PM

Admin
P: 21,827

Somewhere there seems to be an 'R' missing.
Also, one of the latex expression needs \ in front of frac. 


#4
Nov105, 03:49 PM

P: 110

Projectile Motion: Range proof



#5
Nov105, 03:52 PM

HW Helper
P: 876

[tex]\triangle x = Rcos \alpha[/tex]and [tex]\triangle y = Rsin \alpha[/tex] 


#6
Nov105, 03:57 PM

P: 110

[tex] R Sin \alpha = R tan \beta cos \alpha  \frac {R g {cos}^2 \alpha}{2 {V_0}^2 {cos}^2 \beta} [/tex] but i don`t see how this helps? 


#7
Nov205, 01:05 PM

HW Helper
P: 876

You're almost there. But the R in the last term should be R²



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