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 December 16th, 2016, 11:09 AM #1 Member   Joined: May 2016 From: Ireland Posts: 94 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-(1-tan^2 a)x^2 I'm not sure where to go with this, but I was making the equation 35sin a t-4.9t^2=y and x=35cos a t I don't know where to go from here. Last edited by skipjack; December 18th, 2016 at 11:52 AM.
 December 16th, 2016, 12:12 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,548 Thanks: 1259 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 12:14 PM.
 December 18th, 2016, 08:47 AM #3 Member   Joined: May 2016 From: Ireland Posts: 94 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 11:49 AM.
 December 18th, 2016, 11:25 AM #4 Math Team   Joined: Jul 2011 From: Texas Posts: 2,548 Thanks: 1259

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