Posted by Bob on Tuesday, September 7, 2010 at 8:13pm.
I assume she jumps at 45° above horizontal.
Is her initial velocity given by the question?
No, the initial velocity has to be solved for.
ANSWER:
Vo = 9.0 m/s
She lands 0.21 m past the opposite bank
I'm not sure how to get those numbers, though.
How far a long jumper gets obviously depends upon the takeoff speed. Unless you know what that is, you cannot say if she comes up short or not.
I just said the takeoff speed is 9.0 m/s.
I'm not sure either.
I have three unknowns (x, t and u =initial velocity), and two equations:
x=ucos(θ)t
y=usin(θ)t-(1/2)gt²
unless I am missing something.
In any case, if I substitute u=9 m/s, I get the horizontal distance as 10.27, which is 0.27m past the other bank (and not 0.21).
I thought you said that the take-off speed has to be found (unknown).
You do have to solve for the take-off speed, but I have the answers. I just don't know the procedure for solving the problem.
Neither do I when there are three unknowns and two equations. Sorry.
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