# Thread: simple, challenging catapult problem

1. ## simple, challenging catapult problem

Above is a catapult arm with it's rope hugging it. It is at 85degrees with the vertical.

The catapult arm was moved 60degrees to the opposite direction, released, and fired at a range of 35m.
Considering everything else is constant, what is the rate of change(dr/dΘ) of range with respect to how much angle the spoon was moved?

2. ## Suggestion

The equation for the range of an object is:
$r = \frac{v_isin(2\theta_i)}{g}$
where $v_i$ is the initial velocity, $\theta_i$ is the initial angle of trajectory with respect to the horizontal, and $g$ is acceleration due to gravit, 9.81 $m/s^2$. I would suggest finding $\theta_i$ in terms of spoon angle, $\theta_s$ and then take your $\frac{dr}{d\theta_s}$.

3. Originally Posted by zeg
The equation for the range of an object is:
$r = \frac{v_isin(2\theta_i)}{g}$
where $v_i$ is the initial velocity, $\theta_i$ is the initial angle of trajectory with respect to the horizontal, and $g$ is acceleration due to gravit, 9.81 $m/s^2$. I would suggest finding $\theta_i$ in terms of spoon angle, $\theta_s$ and then take your $\frac{dr}{d\theta_s}$.
That's what I thought too.

It is actually more complicated than that as V is not constant. V changes depending on the force the spoon/rope was pulled and you can know how much force is on the rope because it increases as angle gets closer to 0.

I think what complicates this problem for me is that I need to calculate the force exerted by the string on the spoon as it is rotated(by a hand) and released, and I don't know how to do that.