Recent content by kfox1984

i've done it...Thankyou soo much for your help. can I please just ask one last question; finding the highest position i use the fact that v = 0 in the y dirn, is that correct, and i do the same thing pretty much substituting t again?

re arranging i get - u^2 = xthanα - 1/2.g.(x^2/zcos^2α) Which is still incorrect, i really don't know where i am going wrong sorry

i get - Z = usinα.(x/cosα) - 1/2g[x/ucosα]^2 but now i have a cos^2 which I'm not sure how to get rid of?? Thanks

i've made a substitution... x/ucos9α) = t Z = u.sin(α).(x/cos(α)) - 1/2g[(x/ucos(α))^2] is this correct?

sorry that's supposed to be z = u.sin(α) - 1/2gt^2

ok so s = ut + 1/2at^2... x = ucos(α)t , Z = usin(α)t 1/2gt^2 but I'm still unsure about what i am supposed to do next...I already tried rearranging the X one in terms of t then substituting it into the other...which didn't work...if i divide them then i loose the cos(α) using pythagoras was...

I'm sorry but i still really don't understand what i am supposed to do with these.

or x = [(ucos(α) + V/)2] *t and z = [(usin(α) + V)/2] * t

xcos(α) = (U + V)/2 * t , zsin(α) = (u + V)/2 * t

I don't even know if I'm supposed to be using calculus or not - this is supposed to be an as level mechanics problem

I have done that - i'm not sure which equation to use... I know that tx = ty so i have tried rearranging things and substituting. but i still have v's left in everything... i don't understand where the tan has come from... I'm guessing that i have to use v^2 = u^2 + 2as as the final eqn has...

Thanks - I have already done that but still nothing...

i have to answer this question for an assignment that I need to do for mechanics. I am really really stuck - would somebody please mind helping... A ball thrown at an angle α to the horizontal just clears a wall. The horizontal and vertical distances to the top of the wall are X and Z...