CrashOveride said:
I dont see how that follows from what he said ? anywayz
IF you dont do that, so do you just assume the s = ut + 1/2at<sup>2</sup> equation ?
From what i recall, i dont think your allowed to assume anything.
I was giving an example...
In part (a) of the hypothetical question that I made,
we are asked to derive the equation for the horizontal component of the displacement.
In part (b), we are asked to find the maximum height. To find this maximum height we don't actually need to use an expression for horizontal displacement. [In real exams part (a) usually relates to part (b) though] One way of doing part (b) is to use the expression for the vertical component of the velocity and an expresion for vertical component of displacement. Find when the vertical velocity equals zero and that's when it reaches maximum height. Then put t = {that time} into the expression for vertical displacement.
Not the physics formula. What ryan was saying (from what I understand anyway) is that you can use/quote the general formulae, vertical-velocity = -g t + V sin@ and y = -0.5 g t^2 + V t sin@.
But I I were you I would just derive the formulae... I'm not sure if what ryan is saying is correct.
EDIT: on a second look those 2 formulae look very much like the physics formulae anyway... u = V sin@ or V cos@ depending on whether it's the vertical or horizontal part. If my interpretation on what ryan said is true, you still need to write the V and @ part.