Kinematics Problems: Solve for Time, Height & Deceleration | Physics Help

[dajugganaut]

okay, i have a few questions from school i can't solve, please help me...thanks!

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?

 

[dajugganaut]

for the third question, what i did was i thought that this person must be driving away at 9m/s because his average velocity had to be 21m/s ((33+90)/2=21)

 

[Gokul43201]

What have you done so far to try and solve these problems ? Show us what you've tried and where you're stuck, and we will help unstick you.

 

[dajugganaut]

then what i did was i found the time the car would take the travel through the speed trap if it was traveling at an average velocity of 21m/s:
since v=d/t
then t=d/v
t=110/21
t=5.23...

 

[Gokul43201]

for the third question, what i did was i thought that this person must be driving away at 9m/s because his average velocity had to be 21m/s ((33+90)/2=21)

Okay! So, if his speed drops to 9m/s by the second strip, what must his deceleration be ? You know v(i), v(f) and the distance, s. What formula gives you the acceleration from these quantities ?

 

[Gokul43201]

then what i did was i found the time the car would take the travel through the speed trap if it was traveling at an average velocity of 21m/s:
since v=d/t
then t=d/v
t=110/21
t=5.23...

Okay, so far.

 

[dajugganaut]

then I am presuming i would use this equation:

delta(distance)=v(i)*t+0.5*a*t^2

 

[Gokul43201]

then I am presuming i would use this equation:

delta(distance)=v(i)*t+0.5*a*t^2

This is one way to solve the problem.

So, this way, you don't really need the final velocity (9m/s).

OR, having found the final velocity and time, you could use

a = [ v(f) - v(i) ] / t

OR, having found just the final velocity (no need to calculate time)

use 2ad = [ v(f)^2 - v(i)^2 ]