# [SOLVED] Initial height question in vector spaces

• February 1st 2010, 07:10 AM
unifieda
[SOLVED] Initial height question in vector spaces
A stone is thrown from a rooftop at time t = 0. Its position at time t is given by r(t) = (10t,−5t, 6.4−4.9t^2). The origin is at the base of the building, which is on flat ground. Distance is measured in meters and time in seconds.
(a) How high is the rooftop?

I know that the initial velocity r'(t) = 0 at t = 0. But I am unsure on how to solve for the height of the building using parametric equations.
• February 1st 2010, 08:36 AM
Soroban
Hello, unifieda!

This is a strange one . . . an "eyeball" problem
Just look at it . . .nothing fancy is needed!

Quote:

A stone is thrown from a rooftop at time $t = 0.$
Its position at time $t$ is given by: . $r(t) \:=\:(10t,\: -5t,\: 6.4-4.9t^2)$
The origin is at the base of the building, which is on flat ground.
Distance is measured in meters and time in seconds.

(a) How high is the rooftop?

I know that the initial velocity r'(t) = 0 at t = 0. . This is not true!

We are given three functions: . $\begin{Bmatrix}x \;=\;10t \\ y \;=\;-5t \\ z \;=\;6.4-4.9t^2 \end{Bmatrix}$

I would guess the following:

$x$ is the east-west location of the stone.
. . It is moving east at 10 m/sec. **

$y$ is the north-south location of the stone.
. . It is moving south at 5 m/sec. **

$z$ is the height of the stone above the ground.

At $t = 0,\;z = 6.4$
The stone began at 6.4 m above the ground.

. . Therefore, the height of the building is 6.4 meters.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

The stone is thrown horizontally from the building.
Looking down, the scene looks like this:
Code:

          10t     A * - - - - - *         *  θ    |             *    | 5t               *  |                   * B

The stone is thrown from $A$ in the direction of $B.$

Its speed is: . $\sqrt{10^2 + 5^2} \:=\:\sqrt{125} \:\approx\:11.2$ m/sec.

Its direction is: . $\theta \:\approx\;26.7^o$ south of east.

So, you see, the initial velocity was not zero.

• February 1st 2010, 01:08 PM
unifieda
Quote:

Originally Posted by Soroban
Hello, unifieda!

This is a strange one . . . an "eyeball" problem
Just look at it . . .nothing fancy is needed!

We are given three functions: . $\begin{Bmatrix}x \;=\;10t \\ y \;=\;-5t \\ z \;=\;6.4-4.9t^2 \end{Bmatrix}$

I would guess the following:

$x$ is the east-west location of the stone.
. . It is moving east at 10 m/sec. **

$y$ is the north-south location of the stone.
. . It is moving south at 5 m/sec. **

$z$ is the height of the stone above the ground.

At $t = 0,\;z = 6.4$
The stone began at 6.4 m above the ground.

. . Therefore, the height of the building is 6.4 meters.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

The stone is thrown horizontally from the building.
Looking down, the scene looks like this:
Code:

          10t     A * - - - - - *         *  θ    |             *    | 5t               *  |                   * B

The stone is thrown from $A$ in the direction of $B.$

Its speed is: . $\sqrt{10^2 + 5^2} \:=\:\sqrt{125} \:\approx\:11.2$ m/sec.

Its direction is: . $\theta \:\approx\;26.7^o$ south of east.

So, you see, the initial velocity was not zero.

Thank you so much for the help. I cannot believe I did not see this.