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Sx=Uxt Therefore 17.34 = Ux(1.568) Ux=10.93ms-1 Sy=Uyt + 1/2ay(t^2) -1.93=Uy(1.586) - (0.5x9.8x1.586^2) Therefore Uy = 6.55 ms-1 U = √((10.93^2) + (6.55^2)) = 12.75ms-1 angle = tan^-1 (6.55/10.93) = 30º56' Therefore U = 12.75m/s at 30º56' above horizontal

sup

Thanks mate you're a boss

I'm getting a really complicated answer as well. i think it will just be a messy result

can u post a latex solution to it please?

Method 2 but i dont think that works for this question

yea i can do it algebraically but i just want a worked solution to see if it's right

It's in his book on page 31. He equates the terms really quickly but it generally works for roots that are fairly clean. i think this one is a little more complex :O have any idea on the answer?

Hi i was wondering if someone could find the sqaure roots of (12-6i) I usually use Terry Lee's quick method, but this is a bit harder. Thanks for the help