find an expression for x in terms of t
find an expression for x in terms of t
heres a screenshot of it :
http://prntscr.com/m2p1pl
heres a screenshot of it :
http://prntscr.com/m2p1pl
Original post by ilikebigboottys
What forces act horizontally?
Original post by mqb2766
What forces act horizontally?
it doesnt state any horizontal forces. all the required information is within the screenshot
Original post by ilikebigboottys
it doesnt state any horizontal forces. all the required information is within the screenshot
That's correct, gravity acts vertically and no other forces are mentioned. So if there is no force / acceleration, what sort of motion will it be horizontally?
Original post by mqb2766
That's correct, gravity acts vertically and no other forces are mentioned. So if there is no force / acceleration, what sort of motion will it be horizontally?
not sure, wondering if you could help or explain?
Original post by ilikebigboottys
not sure, wondering if you could help or explain?
What is the suvat equation for position when the acceleration is zero?
Original post by mqb2766
What is the suvat equation for position when the acceleration is zero?
s = (u+v)/2 x t
Original post by ilikebigboottys
s = (u+v)/2 x t
You could use that one and reason about what the final velocity is when the acceleration is zero. Probably easier to use the other one which has an acceleration variable in and set it equal to zero?
In either case, what is the initial velocity "u"?
Original post by mqb2766
You could use that one and reason about what the final velocity is when the acceleration is zero. Probably easier to use the other one which has an acceleration variable in and set it equal to zero?
In either case, what is the initial velocity "u"?
In either case, what is the initial velocity "u"?
should be 0?
Original post by ilikebigboottys
should be 0?
No.
What was the initial velocity vertically in part a?
What is the actual initial velocity on the hypotenuse? What is the initial velocity horizontally (use a right angled triangle)?
Original post by mqb2766
No.
What was the initial velocity vertically in part a?
What is the actual initial velocity on the hypotenuse? What is the initial velocity horizontally (use a right angled triangle)?
What was the initial velocity vertically in part a?
What is the actual initial velocity on the hypotenuse? What is the initial velocity horizontally (use a right angled triangle)?
intial velocity was 26ms^-1
Original post by ilikebigboottys
intial velocity was 26ms^-1
So how does that resolve horizontally and vertically?
Its a simple pythagorean triplet, using the information about sin(theta).
Original post by lconlon13
S=26tx5/13
=10t
=10t
how did you get the 5/13
could you explain it please?
Original post by ilikebigboottys
how did you get the 5/13
could you explain it please?
could you explain it please?
Cos(sin^-1(12/13)) or 5 12 13 pythag triple
ohhhhhhhhhhhhhhhhh i finally got in now. thanks guys 
so it's like this:
http://prntscr.com/m2qcro
so it's like this:
http://prntscr.com/m2qcro
Original post by ilikebigboottys
you really need to, and should be able to, write cos(theta) = 5/13 straight away, when given sin(theta) = 12/13 . WITHOUT having to find theta.
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