A stone is thrown at an angle of 36.87 degrees

[devanlevin]

a stone is thrown at an angle of 36.87 degrees from a height of h=8m and falls to the ground 12m from the point of throwing,
given that the acceleration is defined by
$$\vec{a}$$=-$$\hat{x}$$-10$$\hat{y}$$
find:
a the initial velocity of the throw
b the total time
c the maximum height
d the radius of cruvatuture at maximum height
e the radius of curvature when it hits the ground

i have managed to answer all but the last 2

a using the equations for constant acceleration, saying
x(t$$_{final}$$)=12=Vo*cos37-$$\frac{1}{2}$$t$$^{2}$$
y(t$$_{final}$$)=0=8+Vo*sin37-5t$$^{2}$$
2 equations with 2 variables, find that
$$\vec{V}$$o=(9.021,36.87)
tf=$$\sqrt{136/37}$$s

to find the maximum height i again used the equations for
v$$^{2}$$(t)=Vo$$^{2}$$+2$$\vec{a}$$delta$$\vec{r}$$
knowing that at max height, Vy(t)=0
max height=9.465m

problem now is i don't know how to find the radius of curvature at either point
please help

 

[LowlyPion]

Perhaps this example would help?
http://www.coventry.ac.uk/ec/jtm/11/dg11p4.pdf