Of

*course* it's the Math problem from Hell! It's really a Physics problem.

Seriously, conceptually this isn't so bad, there's just a lot to keep track of. The problem winds up being tricky to solve for in the end, but I'll get you set up for it.

First, there are two objects to consider: the diver and the water bucket. Both of these objects need to have the same coordinates at time t = W. So I am going to derive expressions for each object as they move as a function of W.

Over all: I am going to choose an origin (where x and y are both 0) directly under the center of the ferris wheel on the ground. I will choose +x to the left and +y upward.

The bucket of water:

The y coordinate of the water level is easy: y = 8 (ft) is constant. (I'm going to drop the units from now on.) The x coordinate is almost as simple. The bucket starts at x0 = -240 and moves at a constant v = 15 (ft/s), so the x coordinate is x = -240 + 15t. So at t = W: x = -240 + 15W, y = 8.

The diver:

The diver is a bit trickier. The diver starts with rotational motion, then is released to go into free-fall motion. These are different motions. Let's start with the rotational motion.

The ferris wheel has a period of T = 40 s. Thus the angular frequency is $\displaystyle \omega = 2\pi /T=2\pi/40=0.15708$. (For the record this is 0.15708 rad/s.) The diver here is simply moving in a circle with radius 50 (ft), so his speed is a constant $\displaystyle v=r\omega=(50)(0.15708)=7.85398 \, ft/s$. Now, the speed is constant, but we need the components of the diver's velocity in the x and y directions. Again, the diver is simply moving in a circle so this is easy: $\displaystyle v_x=-r\omega cos(\omega t) = -(50)(0.15708)*cos(0.15708*t)$ and $\displaystyle v_y=r\omega sin(\omega t)=(50)(0.15708)*sin(0.15708*t)$. (Be sure your calculator is in radian mode!)

We will also need the x and y coordinates of the diver. This is, again, fairly simple to figure: $\displaystyle x = -50cos(\omega t)$ and $\displaystyle y = 65 + 50 sin( \omega t)$. (Recall the center of the ferris wheel is at (0, 65) and the diver starts at 3 o'clock and thus starts at (-50, 65).)

The diver is released sometime before t = W. Call this time t' for reference.

So when the diver is released, he is at $\displaystyle x = -50*cos(0.15708 t')$ and $\displaystyle y = 65+50*sin(0.15708 t')$ and has velocity components

$\displaystyle v_x=-(50)*(0.15708)*cos(0.15708 t')$

$\displaystyle =-7.85398*cos(0.15708 t')$

and

$\displaystyle v_y=7.85398*sin(0.15708 t')$.

Whew! Take a breather for a minute. We're getting there!

Now for the free-fall motion.

The general formula for motion in the x-direction under constant acceleration is $\displaystyle x=x_0+v_{0x}(t-t_0)+1/2*a_x(t-t_0)^2$. (You don't normally see the t0 term since we usually set that to 0. We can't easily do that here, so I'm not going to try.) Since the diver is in free-fall, ax = 0. We know the position and velocity component when the diver was released (at t = t') so:

$\displaystyle x = (-50*cos(0.15708 t'))$

$\displaystyle -(7.85398*cos(0.15708 t'))*(t - t')$

The y formula works the same way, except now the acceleration is the acceleration due to gravity: ay = -32 ft/s/s. (Note the "-" sign. This is because the acceleration is downward.)

$\displaystyle y=(65+50*sin(0.15708 t'))$

$\displaystyle +(7.85398*sin(0.15708 t'))*(t - t') -16(t-t')^2$

Hold on! We're almost there!

So now we have coordinates for both the bucket and diver as functions of t. Recall that in order for the diver to get into the water these coordinates must be equal. So:

$\displaystyle x = -240 + 15W = (-50*cos(0.15708 t')) $

$\displaystyle - (7.85398*cos(0.15708 t'))*(W - t')$

and

$\displaystyle 8 = (65+50*sin(0.15708 t'))+$

$\displaystyle (7.85398*sin(0.15708 t'))*(W - t') -16(W-t')^2$

This is the Math problem you need to solve. I haven't done it yet. It's going to be tricky since it's a system of equations in two unknowns, t' and W, where the t' appears both inside and outside of a trig function. I'll work at it, but off-hand, I don't think I've ever managed to solve one of these.

**Perhaps one of the Math guru's here will be nice enough to lend a hand at this stage.**
(Hopefully they've read this to the end!

)

Best of luck!

-Dan

PS Just had a thought. If we use a new time coordinate T such that T = W - t' and solve the system for t' and T we can keep the t' strictly inside the trig functions in the y equation. Might be helpful, might not be.