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Horizontal projection means theta=0. That should make things easier for u. And v=sqrt(Vx^2+Vy^2)

Not 100 percent sure but I tried hehe.

Re: MX2 Integration Marathon K I did that and got int(6x^3+6-(1/((x-0.5)^2+0.75))) Final result is (3x^4)/2+6x-(2/sqrt3).tan^-1((2x-1)/sqrt3) That right?

Re: MX2 Integration Marathon let u^3=x, 3u^2du=dx So integral is 3u^2/(u^3/2+u).du take out factor of u on bottom, it cancels with one u on top so u have 3u/(sqrtu+1) let x=sqrtu and divide 3x^2 by (x+1), get -3/3x^2 as remainder, so -1/x^2, and the rest expands to 3x^2-3 So just...

Re: MX2 Integration Marathon I got a final answer of x^3-3x+(1/x)+c, where x=sqrtu. Is that right?

x''=0 x'=15sqrt2 x=15sqrt2t y''=-10 y'=-10t+15sqrt2 y=-5t^2+15sqrt2t When y=-10, t=4.67 (2dp) so time of flight=4.67s When t=4.67, x=99.08 (2dp), so distance travelled is 99.08m I think thats right.