June 20th, 2017, 06:09 AM  #1 
Member Joined: May 2016 From: Ireland Posts: 96 Thanks: 1  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 
June 20th, 2017, 09:37 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,750 Thanks: 1400 
$\Delta x = u\cos{\beta} \cdot t  \dfrac{1}{2}g\sin{\alpha} \cdot t^2$ $\Delta y = u\sin{\beta} \cdot t  \dfrac{1}{2}g\cos{\alpha} \cdot t^2$ $\Delta y = 0 \implies t = \dfrac{2u\sin{\beta}}{g\cos{\alpha}}$ $R = u\cos{\beta} \cdot \dfrac{2u\sin{\beta}}{g\cos{\alpha}}  \dfrac{1}{2}g\sin{\alpha} \cdot \left(\dfrac{2u\sin{\beta}}{g\cos{\alpha}}\right)^ 2 = \dfrac{2u^2\sin{\beta}[\cos{\alpha}\cos{\beta}\sin{\alpha}\sin{\beta}]}{g\cos^2{\alpha}}$ $\dfrac{3R}{5} = u\cos{\beta} \cdot \dfrac{u\sin{\beta}}{g\cos{\alpha}}  \dfrac{1}{2}g\sin{\alpha} \cdot \left(\dfrac{u\sin{\beta}}{g\cos{\alpha}}\right)^2 = \dfrac{u^2\sin{\beta}[2\cos{\alpha}\cos{\beta}\sin{\alpha}\sin{\beta}]}{2g\cos^2{\alpha}}$ $R = \dfrac{5u^2\sin{\beta}[2\cos{\alpha}\cos{\beta}\sin{\alpha}\sin{\beta}]}{6g\cos^2{\alpha}}$ $\dfrac{2u^2\sin{\beta}[\cos{\alpha}\cos{\beta}\sin{\alpha}\sin{\beta}]}{g\cos^2{\alpha}} = \dfrac{5u^2\sin{\beta}[2\cos{\alpha}\cos{\beta}\sin{\alpha}\sin{\beta}]}{6g\cos^2{\alpha}}$ $12\cos{\alpha}\cos{\beta}12\sin{\alpha}\sin{\beta} = 10\cos{\alpha}\cos{\beta}  5\cdot \sin{\alpha}\sin{\beta}$ $2\cos{\alpha}\cos{\beta} = 7\sin{\alpha}\sin{\beta}$ $\dfrac{2}{7} = \dfrac{\sin{\alpha}\sin{\beta}}{\cos{\alpha} \cos{\beta}} = \tan{\alpha}\tan{\beta}$ 
June 21st, 2017, 06:52 AM  #3 
Member Joined: May 2016 From: Ireland Posts: 96 Thanks: 1 
thank you


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