1. ## vectors

1.) For the two vector x1= (20 m) x^ and x2= (15 m) x^, compute and show graphically (a) x1 + x2, (b) x1 - x2, and (c) x2 - x1. [The one and the two after the x's are supposed to be subscripts.]

2.) Given two vectors A, which has a length of 10.0 and makes an angle of 45° below the -x-axis, and B, which has an x-component of +2.0 and y-component of +4.0, (a) sketch the vectors on x-y axes, with all their "tails" starting at the origin, and (b) calculate A + B.

3.) A 2.05 m tall basketball player takes a shot 6.02 meters from the basket (At 3-point line). If the launch is 25° and the ball is launches at the level of the players head, what must the release speed of the ball for the player to make the shot? The basket is 3.05 m above the floor.

2. Hello, John 5!

3.) A 2.05 m tall basketball player takes a shot 6.02 meters from the basket.
If the launch is 25° and the ball is launched at the level of the player's head,
what must the release speed of the ball for the player to make the shot?
The basket is 3.05 m above the floor.

We are expected to know the "trajectory equations":

. . $\displaystyle x \;=\;(v\cos\theta)t$
. . $\displaystyle y \;=\;h_o + (v\sin\theta)t - 4.9t^2$

. . . . where: .$\displaystyle \begin{Bmatrix}v & = & \text{initial speed} \\ \theta & = & \text{angle of release} \\ h_o & = & \text{initial height} \end{Bmatrix}$

We are given: .$\displaystyle \theta = 25^o$
We will let: .$\displaystyle h_o = 0$

Our equations are: .$\displaystyle \begin{array}{cccc}x & = & (v\cos25)t & [1] \\ y & = & (v\sin25)t - 4.9t^2 & [2]\end{array}$

And we want: .$\displaystyle x = 6.02,\;y = 1$

Set [1] equal to 6.02: .$\displaystyle (v\cos25)t \:=\:6.02\quad\Rightarrow\quad t \:=\:\frac{6.02}{v\cos25}$ .[3]

Set [2] equal to 1: .$\displaystyle (v\sin25)t - 4.9t^2 \:=\:1$

Substitute [3]: .$\displaystyle (v\sin25)\left(\frac{6.02}{v\cos25}\right) - 4.9\left(\frac{6.02}{v\cos25}\right)^2 \;=\;1$

We have: .$\displaystyle 6.02\tan25 - \frac{4.09(6.02^2)}{v^2\cos^2\!25} \;=\;1\quad\Rightarrow\quad\frac{4.9(6.02^2)}{v^2\ cos^2\!25} \;=\;6.02\tan25 - 1$

Then: .$\displaystyle \frac{v^2\cos^2\!25}{4.9(6.02^2)} \;=\;\frac{1}{6.02\tan25 - 1}\quad\Rightarrow\quad v^2\;=\;\frac{4.9(6.02^2)}{\cos^2\!25(6.02\tan25 - 1)}$

Hence: .$\displaystyle v \;=\;\frac{6.02\sqrt{4.9}}{\cos25\sqrt{6.02\tan25 - 1}}$

. . And I got: .$\displaystyle v\:\approx\:10.94$ m/sec

A practical note:
This speed allows the ball to meet the rim of the basket.
It does not guarentee a score.

3. Hmm couldn't you solve number 3 with vectors? I've yet to learn about them a lot and I would find it neat to see how.

4. Originally Posted by Jonboy
Hmm couldn't you solve number 3 with vectors? I've yet to learn about them a lot and I would find it neat to see how.
Soroban did do the problem using vectors. The notation is such that the vector nature is buried. x and y are components of the displacement vector, just as $\displaystyle v_x$ and $\displaystyle v_y$ are components of the velocity.

-Dan

5. Thanks Soroban!!!! I would have never known how to even start the problem.

Any idea on the first two problems?

6. Hello again, John 5!

2.) Given two vectors:
$\displaystyle \vec{A}$, which has a length of 10.0 and makes an angle of 45° below the x-axis,
and $\displaystyle \vec{B}$, which has an x-component of +2.0 and y-component of +4.0.

(a) Sketch the vectors on x-y axes, with all their "tails" starting at the origin.

(b) Calculate $\displaystyle \vec{A} + \vec{B}$
Code:
      |     B
|   * (2,4)
|  /:
| / :
|/  :
O * - + - - + - -
| \ 45°   :
|   \     :
|  10 \   :
|       \ :
|         * A

We have: .$\displaystyle \vec{A} \;=\;\overrightarrow{OA} \;=\;\langle5\sqrt{2},\,\text{-}5\sqrt{2}\rangle$

. . , ,and: .$\displaystyle \vec{B} \;=\;\overrightarrow{OB} \;=\;\langle 2,\,4\rangle$

Therefore: .$\displaystyle A + B \;=\;\langle2 + 5\sqrt{2},\:4-5\sqrt{2}\rangle$