Question

A diver springs upward from a board that is 3.10 meters above the water. At the instant she contacts the water her speed is 9.07 m/s and her body makes an angle of 70.5° with respect to the horizontal surface of the water. Determine her initial velocity.

Answer 1

the final velocity vector tells us that her final horizontal velocity =9.07 cos 70.5 = 3.02m/s

since there are no horizontal forces acting on her, this was her horizontal velocity throughout the trip, and the horizontal velocity she left the board with

her final vertical velocity = 9.07sin 70.5 = 8.54m/s

we can find her initial vertical velocity using

vf^2=v0^2 + 2ad
vf=9.07m/s
v0=unknown
a=9.8m/s/s
d=3.1m

v0^2 = 9.07^2 -2*9.8*3.1 = 3.18m/s

therefore, the speed at launch = sqrt[3.18^2+2.08^2]