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June 20th, 2017, 06:09 AM   #1
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projectile fired

a plane is inclined at an angle a to the horizontal . a particle up the plane with speed u at an angle b to the plane . when the particle is at its max perpendicular height above the plane it is 3/5 of the range up the plane. show tana*tanb=2/7

so far I have found the time of flight which is 2usinb/gcosa
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June 20th, 2017, 09:37 AM   #2
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$\Delta x = u\cos{\beta} \cdot t - \dfrac{1}{2}g\sin{\alpha} \cdot t^2$

$\Delta y = u\sin{\beta} \cdot t - \dfrac{1}{2}g\cos{\alpha} \cdot t^2$

$\Delta y = 0 \implies t = \dfrac{2u\sin{\beta}}{g\cos{\alpha}}$

$R = u\cos{\beta} \cdot \dfrac{2u\sin{\beta}}{g\cos{\alpha}} - \dfrac{1}{2}g\sin{\alpha} \cdot \left(\dfrac{2u\sin{\beta}}{g\cos{\alpha}}\right)^ 2 = \dfrac{2u^2\sin{\beta}[\cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta}]}{g\cos^2{\alpha}}$

$\dfrac{3R}{5} = u\cos{\beta} \cdot \dfrac{u\sin{\beta}}{g\cos{\alpha}} - \dfrac{1}{2}g\sin{\alpha} \cdot \left(\dfrac{u\sin{\beta}}{g\cos{\alpha}}\right)^2 = \dfrac{u^2\sin{\beta}[2\cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta}]}{2g\cos^2{\alpha}}$

$R = \dfrac{5u^2\sin{\beta}[2\cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta}]}{6g\cos^2{\alpha}}$

$\dfrac{2u^2\sin{\beta}[\cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta}]}{g\cos^2{\alpha}} = \dfrac{5u^2\sin{\beta}[2\cos{\alpha}\cos{\beta}-\sin{\alpha}\sin{\beta}]}{6g\cos^2{\alpha}}$

$12\cos{\alpha}\cos{\beta}-12\sin{\alpha}\sin{\beta} = 10\cos{\alpha}\cos{\beta} - 5\cdot \sin{\alpha}\sin{\beta}$

$2\cos{\alpha}\cos{\beta} = 7\sin{\alpha}\sin{\beta}$

$\dfrac{2}{7} = \dfrac{\sin{\alpha}\sin{\beta}}{\cos{\alpha} \cos{\beta}} = \tan{\alpha}\tan{\beta}$
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June 21st, 2017, 06:52 AM   #3
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thank you
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