ap physics please helpp pleasee???

Question

A diver springs upward from a board that is three meters above the water. At the instant she contacts the water her speed is 9.10 m/s and her body makes an angle of 73.0° with respect to the horizontal surface of the water. Determine her initial velocity, both magnitude and direction.
___m/s (magnitude)
____° (direction)

Answer 1

Assume the horizontal component of velocity was constant,
so 9.10*cos(73)=vx
vx is constant

That means that the vertical at impact with the water is
9.10*sin(73)

This is related to the distance from apogee.
m*g*h=.5*m*v^2 where h is the height of apogee above the water
vy^2 at impact=2*g*h

solve for h
h=((9.10*sin(73))^2)/(2*g)

h is the apogee, and the starting point is h-3

using
y(t)=3+vy0*t-.5*g*t^2
and
vy(t)=vy0-g*t
when vy(t)=0, y(t)=h
vy0/g=t
plug into y(t)=h
h=3+vy0^2/g-.5*vy0^2/g
h-3=.5*vy0^2/g
vy0=sqrt(2*g*(h-3))

we know h, solve for vy0
The magnitude of the take-off is
sqrt(vy0^2+vx^2)
the direction above horizontal is
ATAN(vy0/vx)


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Answer 2

minimum on the best of the arc, optimal on the backside (the two ends) of the arc. Acceleration through gravity is the reason. because of the fact the ball is going up, its speed decreases because of the stress of gravity until eventually its effective speed is 0 (vertically). because it is going backpedal it useful properties acceleration until eventually it contacts the floor. provided the landing element is on a similar top because of the fact the commencing element, the two ends would have a similar (opposite) speed. Horizontal speed continues to be consistent throughout the test as long as we are discounting air resistance, subsequently its result on the relative speed is skipped over.