1. Sep 13, 2004

### Doug Desatnik

Hi Everybody,

I am having a problem with a homework question and think that I have finally got it right this time! Can somebody verify my work on this problem and let me know if I did indeed do this problem correctly? If not then can anybody help point me in the right direction and get me back on track?

The HW deals with Kinematics in 2-Dimensions

THANK YOU!!!

- Doug

--------------------------------------------------------------------------
The Question
A rock climber throws a small first aid kit to another climber who is higher up the mountain. The initial velocity of the kit is 6.93 m/s at an angle of 66.5 degrees above the horizontal. At the instant when the kit is caught, it is traveling horizontally, so its vertical speed is zero. What is the vertical height between the two climbers?

My Set-Up of Variables

V_initial = 6.93 m/s
V_final - 0 m/s
A = 9.80 m/s^2
Y = ? (SOLVE FOR)
T = ?

My Equation & Work

V^2=V_initial^2+2AY --> Y = (V_final^2 - V_initial^2)/2A --> (0^2 - 6.93^2)/2(9.80) --> 48.02/19.6 --> 2.45m

2. Sep 13, 2004

### Doug Desatnik

I wanted to try this question using this LATEX code, so here it is:

My Variables
$$V{0}$$ = 6.93 m/s
V = 0 m/s
a = $$9.80 m/s^2$$

My Equation
$$V^2=V{o}^2+2aY$$

My Work

$$V^2=V{o}^2+2aY$$ --> $$Y=V^2 - V{o}^2/2a$$ --> $$Y=(0^2 - 6.93^2)/(2)(9.80)$$ --> $$2.45m$$

LATEX is really cool! It will surely take some time to get used to but this is really a nice function!!!

3. Sep 13, 2004

### Doug Desatnik

Ok, so after looking at this problem a bit differently, it is obvious that what the question is asking for is the max height of the projectile, right? In that case then the equation would be:

$$V^2=V{o}^2-2gy$$

$$V{o}=6.93 m/s$$
$$V=0 m/s$$
$$g=9.80 m/s^2$$
$$y=?$$

I guess then my question becomes a simply math (algebra) problem..... How do I rearrange the equation above to solve for the height, Y?

Is it like so?

$$Y=(V^2-V{0}^2)/-2g$$

- Doug