- #1
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
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
