1. Jul 7, 2004

### dajugganaut

1. A Pentium 1 computer is discovered in a room, which is 4.5m above the ground. A person walks by and chucks the computer out the room window with an upwards velocity of 6.0m/s. How long does it take before the computer hits the ground?

2. To help pay his bills, Bill works as a roofer, laying down one shingle after another. One day, Bill dropped a roof tile off the top of a building. An observer with a stopwatch inside the bulding notics that it takes 0.20s for the tile to pass his/her window, whose height is 1.6m. How far above the top of this window is the roof?

3. While Bill is going through his driving test, he unfortunately passes over a speed trap which is set up with 2 pressure-activated strips placed across the road, 110m apart. Bill was speeding along at 33m/s while the speed limit was 21 m/s. At the instant the car activates the first strip, Bill begins slowing down. What minimum deceleration is needed in order that the average speed is within the limit by the time the car crosses the second marker?

2. Jul 7, 2004

### AKG

Question 1:

$\Delta \vec{d} = 4.5m\ \mbox{[down]}$
$\vec{v}_1 = 6.0m/s\ \mbox{[up]}$
$\vec{a} = 9.8m/s^2\ \mbox{[down]}$
$\Delta t = ?$

Question 2

You have 3 knowns, and would like to know the initial velocity of the brick as it passes the top of the window. You can do this. Now, to find the distance from the roof to the top of the window, you have three knowns (initially, the velocity is zero, since it's dropped, you have acceleration, and you have final velocity, which is the velocity of the brick as it just passes the top of the window calculated previously). Solve for displacement.

Question 3

The average speed is the total distance over the total time. You know the total distance, and the speed he's allowed to have, so you can figure out the time he must take. You also know his initial velocity. 3 knowns, solve for the unknown, acceleration.

3. Jul 7, 2004

question 1: d = (vo)t + 1/2at^2 {vo = initial velocity}
sub in d as -4.5m
sub in vo as 6.0m/s
and a is -9.8m/s^2

question 2:
first u need the initial veocity of the projectile as it passes the very top of the window
you do this by using the equation d = vot + 1/2at^2
sub in d = 1.6m
sub in t = 0.2s
sub in a = 9.8m/s^2
now solve for vo, Im guessing you know how to do this.
after you have found vo, then you know that the rock is dropped with no initial velocity, so you use the equation V^2 = vo^2 + 2ad
sub in V = previous calculated answer (vo)
sub a = 9.8m/s^2
sub in vo = 0m/s since the rock is dropped with no initial velocity
now all you do is sove for d.

question 3:
first for this question, what we should do is calculate the maximum speed needed at the end of the rope in order to be within the limit. To do this we can use the equation:
Vav = (Vo + Vf)/2
sub in Vav as 21m/s
sub in Vo as 33m/s
sove for Vf or Vfinal (you should get 9m/s)
now that you have final speed you can sub that speed and the known variables into:
sub in Vf = 9m/s
sub in Vo = 33m/s
sub in d as 110m
sove for a, you should get a negative answer.

I hope I've been some help.

4. Jul 7, 2004