# Thread: Vector-valued functions and motions in space

1. ## Vector-valued functions and motions in space

A golf ball is hit with an initial speed of 116ft/sec at an angle of elevation of 45 degrees from the tee to a green that is elevated 45ft above the tee as shown in the diagram. Assuming that the pin, 369ft downrange, does not get in the way, where will the ball land in relation to the pin?

I thought I knew how to do this problem my answer differs from the book's. Here's mine:

Find time (t)
$\displaystyle x= (v_{0}cos\alpha)t \Rightarrow 369=(116cos45\deg)t \Rightarrow t=\frac{369}{(58\sqrt{2})}\approx 4.5$

Then to find y I did:

$\displaystyle y= y_{0}+(v_{0}sin\alpha)t-\frac{1}{2}gt^2\Rightarrow y=-45+(116sin45\deg )4.5-\frac{1}{2}(32)(4.5)^2$
$\displaystyle \Rightarrow -45+369.11 -324=0.11$

The problem is that for the answer in the book they don't use -45 at all and their final answer is that the ball is 45.11 feet in the air when it reaches the pin 369 feet away and therefore goes past the pin.

I don't understand why the fact that the tee is 45 feet lower than the green doesn't factor into the answer. Can someone explain this to me?

2. "$\displaystyle y_0$" is the initial height of the ball. Since the green is 45 feet higher than the ball, $\displaystyle y_0$ is +45, not -45.

3. y0 should be 0 otherwise if y0 = 45 y would be 90.11 ft when it passes over the pin

4. Thanks. I realized my error this morning while brushing my teeth. The golfer is standing on on the y-axis. Therefore, $\displaystyle y_{0}$ is 0. It didn't make sense earlier but I guess taking a break from it for a while cleared things up.