# Kinematics Last minute homework help

1. Sep 3, 2007

### koolkris623

1. The problem statement, all variables and given/known data
Hi I seriously need help in any of these remaining kinematics problems...I need to understand how to solve them too...Any help will be greatly appreciated.

1.) A bird watcher meanders through the woods, walking 0.40 km due east, 0.70 km due south, and 2.20 km in a direction 45.0° north of west. The time required for this trip is 2.50 h. Determine the magnitude and direction (relative to due west) of the bird watcher's

(a) displacement and
km (magnitude)
° (direction north of west)
(b) average velocity.
km/h (magnitude)
° (direction north of west)

Use kilometers and hours for distance and time, respectively

8.) A car drives straight off the edge of a cliff that is 50 m high. The police at the scene of the accident note that the point of impact is 140 m from the base of the cliff. How fast was the car traveling when it went over the cliff?
______(m/s)

16.) A soccer player kicks the ball toward a goal that is 29.0 m in front of him. The ball leaves his foot at a speed of 18.5 m/s and an angle of 33.6° above the ground. Find the speed of the ball when the goalie catches it in front of the net. (Note: The answer is not 18.5 m/s.)
________________m/s

2. Relevant equations
x(t) = .5at2 + v0t + x0
x(t) = .5(v0+vf)t
vf^2= v^2+ 2ax
v= v0+ at

3. The attempt at a solution
I seriously don't know how to do these remaining problems

Last edited: Sep 3, 2007
2. Sep 3, 2007

### cepheid

Staff Emeritus
1a) Tell me, what is the definition of displacement? Based on that answer, how would you compute it?

3. Sep 3, 2007

### Staff: Mentor

This might help.

http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html
Horizontal launch - http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra11
Launch at an angle - http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra3

If a mass is launched at angle $\theta$ with respect to horizontal, then vx0 = v0 $cos \theta$, and vy0 = v0 $sin \theta$

Please show some effort on one's part.

4. Sep 3, 2007

### koolkris623

displacement is the distance basically..it is x in the equations

5. Sep 3, 2007

### cepheid

Staff Emeritus
Notice you're dealing with a problem in two dimensions. There's a key difference between displacement and distance...

6. Sep 3, 2007

7. Sep 3, 2007

### cepheid

Staff Emeritus
uh, anyway, keep us posted with how you're doing with these problems...

8. Sep 4, 2007

### frasifrasi

here is an idea: why not just solve the problem and get it over with--that is why we come here in the first place, because we need to know how to get the answer(so we can learn from it), not to get a 4 hr/long distance/useless tutoring section.

9. Sep 4, 2007

### cristo

Staff Emeritus
So you want someone to do your homework for you? Well, I'm sorry, that's not what we're here for! You should re-read the PF guidelines to which you ageed when you joined. Besides, the majority of students that seek help here find the tutoring approach far more useful than simply providing solutions.