December 16th, 2016, 10:09 AM  #1 
Member Joined: May 2016 From: Ireland Posts: 96 Thanks: 1  projectile fired
A particle is projected from a point P on a horizontal plane with initial speed 35 m/s at an angle a to the plane. Show that if x and y are the horizontal and vertical distances of the particle from P then 250y=250(tan a)x(1tan^2 a)x^2 I'm not sure where to go with this, but I was making the equation 35sin a t4.9t^2=y and x=35cos a t I don't know where to go from here. Last edited by skipjack; December 18th, 2016 at 10:52 AM. 
December 16th, 2016, 11:12 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,737 Thanks: 1387 
your resultant equation should be $250y=250(\tan{a})x(1 \color{red}{+} \tan^2{a})x^2$ $x=v_0\cos{a} \cdot t \implies t = \dfrac{x}{v_0\cos{a}}$ substitute $\dfrac{x}{v_0\cos{a}}$ for $t$ in the equation ... $y = v_0\sin{a} \cdot t  \dfrac{1}{2}gt^2$ ... to eliminate the parameter, $t$. Last edited by skeeter; December 16th, 2016 at 11:14 AM. 
December 18th, 2016, 07:47 AM  #3 
Member Joined: May 2016 From: Ireland Posts: 96 Thanks: 1 
When I sustitute t in, I get xtana (gx^2)/(2u^2cos^2a) how do you go from here to the answer and you were right about the +
Last edited by skipjack; December 18th, 2016 at 10:49 AM. 
December 18th, 2016, 10:25 AM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,737 Thanks: 1387  

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