draw her position just short of the take-off from the ramp.
Let her height be y m and her horizontal distance x m.
By similar triangles, y/x = 5/1
x = y/5
let her distance up the ramp be d
then d^2 = x^2 + y^2
d^2 = y^2/25 + y^2 = (26/25)y^2
d = (√26/5)y
dd/dt = (√26/5)dy/dt
(notice we don't need dx/dt)
I will assume that the 12 m/s is her speed as she goes up the ramp, so ...
12 = (√26/5)dy/dt
dy/dt = 60/√26 m/x
So as long as she is on the ramp, she has a vertical speed of 60/√26 m/s.
So at the moment she leaves the ramp that will be her vertical speed.
Are you sure you don't have your vertical and horizontal numbers mixed up?
That ramp looks extremely unreasonable, she will practically crash into it.
A water skier skis over the ramp, 5m vertical, 1m horizontal, at a speed of 12m/s. How fast is she risingas she leaves the ramp?
d=root26
dy/dt=[(d)(d')-(x)(x')]/y
dy/dt=[(12)(root26)-(5)(x')]/1
How do I find dx/dt?
Thanks in advance.
2 answers
Whoops, I do have the horizontal and vertical numbers mixed up. The horizontal side of the ramp is 5m, and vertical 1m. So I just have to switch the 5 and 1 around right? Thank you so much :)
0 answers