Posted by sara on Wednesday, September 22, 2010 at 8:33am.
Let the angle with the horizontal be φ.
and u the take-off speed (m/s)
t = airborne duration (seconds)
The vertical initial velocity, uy
= u sin(φ)
Horizontal velocity, ux
= u cos(φ)
When she lands, vertical distance=0:
0=uy t - (1/2)gt²
t=2(uy)/g
=2u sin(φ)/g
Horizontal distance (with constant velocity ux)
= ux t
= u cos(φ) * 2u sin(φ)/g
= u²sin(2φ)/g
Can you take it from here?
Yes. I can handle it from here. I just needed a place to start. Thanks MathMate!
You're welcome!
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