# Kinematics equation which is Y = vt+ 1/2gt^2

1. Sep 11, 2008

### kmikias

Hello guys....Here again i came up with another question I guess you help me out.Here is the question.

1.A rock climber tosses a small first aid kit to another climber higher on the mountain.the kit is thrown at velocity of 11m/s at velocity of 65 degree from horizontal.when the kit is caught,it is at the apex of its path.how high up is the second climber?

Here is what i did.

I separate it in to two which is

Horizontal
X = vt
V = 11m/s
V(x) = (cos 65) * 11
X(final)=

Vertical
Y = vt + 1/2gt^2
v=11
v(y)=(sin65)11
t=
y(final)=

I have all given value but how can i find the time with out knowing at list Y final or

Y final with out knowing the time.

i can plug in the formula but i have to have either time or y final.

thank you

2. Sep 11, 2008

### BuGzlToOnl

Re: problem

Haha I did this problem yesterday for an online quiz, I'll help you out a bit.

This is how I did it. The initial velociy is 11 m/s.

V_0 = 11 m/s

The following equation will give you your x-component of velocity and your y-component of velocity respectively. Which is what you said you needed.

v_0x = v_0(cos65)
v_0y = v_0(sin65)

You know the acceleration of a_y = -9.8 m/s2.

And use one of your kinematic equations to solve the problem

After you get your final velocity (use one of your kinematic equations its simple). The formula for height is (v_y)^2/(2)(a_y).

I edited the original content b/c I think I was giving too much away, plus I already have a warning for when I did my first post.

Hope this isn't TOO much help.

Last edited: Sep 11, 2008
3. Sep 11, 2008

### kmikias

Re: problem

thanks for your help.I got the answer finally using V(final) = v(inital) + gt ..........its vertical
v final = 0
so t= 1.017.
After that i use kinematics equation which is Y = vt+ 1/2gt^2.

how ever i have a question for you.

how did you drive the Hight equation

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