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
[tex]\vec{a}[/tex]=-[tex]\hat{x}[/tex]-10[tex]\hat{y}[/tex]
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[tex]_{final}[/tex])=12=Vo*cos37-[tex]\frac{1}{2}[/tex]t[tex]^{2}[/tex]
y(t[tex]_{final}[/tex])=0=8+Vo*sin37-5t[tex]^{2}[/tex]
2 equations with 2 variables, find that
[tex]\vec{V}[/tex]o=(9.021,36.87)
tf=[tex]\sqrt{136/37}[/tex]s

to find the maximum height i again used the equations for
v[tex]^{2}[/tex](t)=Vo[tex]^{2}[/tex]+2[tex]\vec{a}[/tex]delta[tex]\vec{r}[/tex]
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

 

[thomate1]

I would respond by saying that the radius of curvature at maximum height and when the stone hits the ground can be calculated using the formula R = v^2/a, where v is the velocity and a is the acceleration. At maximum height, the velocity is only in the x-direction, so the radius of curvature would be infinite. When the stone hits the ground, the velocity is in both the x and y directions, so the radius of curvature would be R = (v_x^2 + v_y^2)/a. Using the values for v and a calculated in the previous steps, you can determine the radius of curvature at both points.