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Quoted from vlad_drac

I know what you mean, yeah. But do you know what i mean? :/

If he’s throwing it at an angle upwards, we don’t know the initial horizontal or vertical speed. We don’t know the time, we only know the horizontal distance (for uneven ground) and the vertical acceleration.

If you do a quick diagram yourself, i think you might see what i mean. Whether it’s vcos34 or vsign34, it’s still a proportion of a variable that we don’t know.

So basically, we know we know horizontal accel is 0, horizontal displacement is 30.

But if we know what vcos34= and what vsin34= by splitting the parabolic motion and calculating horizontal and vertical, then we can work out v by using simultaneous equations, showing us the unknown proportion. Now i really need to sleep, If I get time when i wake up, ill grab some paper, as i have a notion that I’m not really coming across with what i actually mean :P

I imagine i’ll remember how to do it soon enough, as it looks very similar to the kind of thing I would have tackled a few years ago in Further mechanics (if its solveable ofc)

Hey guys this thread at the moment is too clever for http://www.etf2l.org to handle. So I will just leave this post here.

Quoted from Chaplain

[…]

lolthx Chris. I’ve written the Horizontal Accel as 0, not as 1, so i’ll re do it from there

now we use S=ut+1/2at^2 again

S=xt+1/2t^2 (as a=1)

30=2.47x + 1/2*2.47^2

x=10.91m/s

x(Initial Speed) to 2sf=11m/s

Lol thanks Chris, epic fail on my part there

did i just get trolled? :(

Given myself a migraine by stupidly spending 7 hours on and off staring at this trying to work it out. Taking a break from it today. Not saying anyone’s right or wrong, but so far i’ve either not been able to follow what people were saying (either cos they’re not showing their working or because it’s just hard via an internet forum) or they’ve got a different answer than the stated one.

I didn’t go out through all the replies, but since this doesn’t end with a post of OP being the epitome of thankfulness, I supose there isn’t a solution yet?

We have: y(t0)=0, x(t0)=0, y(te)=-30, x(te)=30 (this only matters because of signs)

Denote the initial velocity as v0. That’s what we are looking for. It has an x- and an y-component, for which we derive the motion equations. Since there is no friction we obtain:

dotx(t)=cos(34°)*v0 (constant)
doty(t)=sin(34°)*v0-g*t (g*t is negative because the ball is initially thrown upward)

We integrate this, thus obtaining position equations & fill in the initial and final condition (note that we have chosen x0=y0=0):

30=v0*te*cos(34°)
-30=v0*te*sin(34°)-0.5*g*te^2

Two equations, two unknowns :)