# What is the vertical height.

1. Sep 3, 2007

### afcwestwarrior

1. The problem statement, all variables and given/known dataA rock climber throws a small first aid kit to another climber who is higher up the mountain. The initial velocity of the kit is 11 m/s at an angle of 65 degrees above the horizontal. At the instant when the kit is caught, it is traveling horizontally, so it's vertical speed is zero. What is the vertical height between the 2 climbers.

Initial Velocity=11 m/s
Angle above horizontal=65 degrees
Vertical speed=0

2. Relevant equations
equations of kinematics

3. The attempt at a solution

2. Sep 3, 2007

### Staff: Mentor

Please show some work and effort on one's part.

Initial Velocity=11 m/s

Determine the horizontal and vertical velocity components.

The horizontal velocity is constant if one ignores air resistance. The kit decelerates as it travels vertically.

Refer to -
http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra5

3. Sep 3, 2007

### afcwestwarrior

Here is what I did. I used the equation y=h=Vy^2 - Voy^2/ 2 (Ay)

Then I used the values I had h=(0m/s)^2 - (11 m/s)^2/ 2 (-9.8m/s^2) = 6.1 meters

The answer is 5.1 meters what did I do wrong

4. Sep 3, 2007

### Staff: Mentor

vy0 is not 11 m/s.

v0 = 11 m/s.

Resolve v0 into vy0 and vx0

Think about sin and cos of the angle with respect to horizontal.

5. Sep 3, 2007

### afcwestwarrior

ok Vy0=11 sin 65=10 m/s
and Vx0=11 cos 65=4.6 m/s

Now what do I do

6. Sep 3, 2007

### afcwestwarrior

Ok so it's h=0 m/s)^2 - 10 m/s) ^2/2 *-9.8 m/s^2=5.1

7. Sep 3, 2007

### afcwestwarrior

Did I do it right

8. Sep 3, 2007

Yes!

9. Sep 3, 2007

### afcwestwarrior

Thank you Astronuc. It feels good learning how to solve a problem. Instead of just writing down the answer.

10. Jan 30, 2008

### mr_schticker

Thanks!

Thanks so much, I had the same problem. Ironically, I had split it up into components already, just for some reason I hadn't added it in to the formula as my VoY, I was still using the old Vo vector.

Thanks!