- #1
zanazzi78
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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.
[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.
Last edited: