Math help - Printable Version +- Southperry.net (https://www.southperry.net) +-- Forum: Social (https://www.southperry.net/forumdisplay.php?fid=14) +--- Forum: Rubik's Cube (https://www.southperry.net/forumdisplay.php?fid=58) +--- Thread: Math help (/showthread.php?tid=34917)

Math help - 2147483647 - 2010-12-22 Assuming parabolic motion, the horizontal velocity vector of a projectile is v(0) cos(θ. The vertical velocity vector is modeled by v(0) sin(θ t + gt. Therefore, we integrate to find the positions: ∫ v(t) dt = ∫ v(0) cos(θ dt = x(t) v(0) cos(θ t +x(0) = x(t) ∫ v(t) dt = ∫ v(0) sin(θ +gt dt = y(t) v(0) sin(θ t +gt^2/2 +r(0) = y(t) Since x(0) is 0 at my starting point, I'm going to rewrite this as: v(0) cos(θ t = x(t) I'm trying to find a general solution that allows for me to clear a wall of height h distance d away. Thus, x(t) = d and y(t) = h. Therefore, my problem is to eliminate t. If I rearrange x(t), I get: t = d /(v(0) cos(θ) Subbing into y(t), I get: g(d /(v(0) cos(θ))^2/2 +v(0) sin(θ (d /(v(0) cos(θ)) +r(0) = h g(d /(v(0) cos(θ))^2/2 +v(0) sin(θ (d /(v(0) cos(θ)) +r(0) -h = 0 Because I have to clear the wall: g (d /(v(0) cos(θ) )^2/2 +v(0) sin(θ (d /(v(0) cos(θ) ) +r(0) -h > 0 Simplifying: gd^2 (sec(θ)^2 /(2 v(0)^2) + d tan(θ + r(0) -h > 0 gd^2 (1+(tan(θ)^2) /(2 v(0)^2) + d tan(θ + r(0) -h > 0 gd^2 (tan(θ)^2 /(2 v(0)^2) + d tan(θ + r(0) -h + gd^2 /(2v(0)^2) > 0 Now I can use the quadratic formula to solve for tan(θ. tan(θ > ( -d +/- √(d^2 -4(gd^2/(2v(0)^2)(gd^2 /(2v(0)^2) +r(0) -h)) ) /(gd^2 /v(0)^2) tan(θ > ( -d +/- √(d^2(1 -(2g/v(0)^2)(gd^2 /(2v(0)^2) +r(0) -h)) ) /(gd^2 /v(0)^2) tan(θ > ( -d +/- d√((1 -(2g/v(0)^2)(gd^2 /(2v(0)^2) +r(0) -h)) ) /(gd^2 /v(0)^2) tan(θ > (v(0)^2) ( -1 +/- √((1 -(2g/v(0)^2)(gd^2 /(2v(0)^2) +r(0) -h)) ) /(gd) tan(θ > (-v(0)^2 +/- v(0)√((v(0)^2 -(2g)(gd^2 /(2v(0)^2) +r(0) -h)) ) /(gd) tan(θ > (-v(0)^2 +/- v(0)√((v(0)^2 -(gd/v(0))^2 -(2g)(r(0) -h)) ) /(gd) θ > arctan((-v(0)^2 +/- v(0)√((v(0)^2 -(gd/v(0))^2 -(2g)(r(0) -h)) ) /(gd) When I put this into my calculator, however, it is giving me incorrect solutions. I don't know what I did wrong. Math help - Russt - 2010-12-22 v(0) is known? Also, when you use the quadratic formula with an inequality, I don't think it works the way you have it. For example: http://www.wolframalpha.com/input/?i=solve+-x^2+%2B+x+%2B+2+%3E+0+for+x So you want tan(θ to be between the two roots you get. Math help - 2147483647 - 2010-12-22 The only two variables that are unknown are r(0) and θ. Everything else should be known: d, h, v(0), g, etc. (And g is already negative, so y(t) = v(0) cos(θ +gt.) And v(0) < h. Sorry I didn't specify that.