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Projectile Motion: Range proof

by zanazzi78
Tags: motion, projectile, proof, range
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zanazzi78
#1
Nov1-05, 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.
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Astronuc
#2
Nov1-05, 03:24 PM
Admin
Astronuc's Avatar
P: 21,827
Somewhere there seems to be an 'R' missing.

Also, one of the latex expression needs \ in front of frac.
daniel_i_l
#3
Nov1-05, 03:44 PM
PF Gold
daniel_i_l's Avatar
P: 867
delta x = cos(alpha) ??

zanazzi78
#4
Nov1-05, 03:49 PM
P: 110
Projectile Motion: Range proof

Quote Quote by daniel_i_l
delta x = cos(alpha) ??
alpha is the angle of the inclinded plane. it forms a right triangle with the horizontal and the point at which the projectile meets the inclinded plane. therefore delta x is equal to cos alpha.
Fermat
#5
Nov1-05, 03:52 PM
HW Helper
P: 876
Quote Quote by zanazzi78
...
[tex]\triangle x = cos \alpha[/tex]and [tex]\triangle y = sin \alpha[/tex]
...
This is where the R is missing. They should be,

[tex]\triangle x = Rcos \alpha[/tex]and [tex]\triangle y = Rsin \alpha[/tex]
zanazzi78
#6
Nov1-05, 03:57 PM
P: 110
Quote Quote by Fermat
This is where the R is missing. They should be,
[tex]\triangle x = Rcos \alpha[/tex]and [tex]\triangle y = Rsin \alpha[/tex]
ok so i`ve put the 'R' s in and get

[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?
Fermat
#7
Nov2-05, 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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