# A stone is thrown vertically upward.

1. Jan 22, 2014

### PHYSStudent098

1. The problem statement, all variables and given/known data

A stone is thrown vertically upward. On its way up it passes point A with speed v, and point B, 3.00 m higher than A, with speed v/2. Calculate (a) the speed v and (b) the maximum height reached by the stone above the point B.

2. Relevant equations

Kinematic Equations

v = v(knot) + a(t)

V^2 = v(knot)^2 + 2a (x - x(knot))

x - x(knot) = v(knot)t + 0.5a(t)^2

3. The attempt at a solution

Because I'm not given time in the problem statement, I believe I start with Kinematic eq two. If I want to find speed at point A, I don't know how to represent that as an x value. I know "a" is "-9.8" but am not given v(knot)^2. I'm left with too many variables.

Thank you.

2. Jan 22, 2014

### grzz

You have chosen the correct formula.

Use that formula from initial speed v to speed v/2.

3. Jan 22, 2014

### PHYSStudent098

But what is my v(knot) and x(knot)? Thank you.

4. Jan 22, 2014

### grzz

You take point A as the zero reference for distance.

5. Jan 22, 2014

### grzz

BTW does not v$_{o}$ look better?

6. Jan 22, 2014

### PHYSStudent098

V^2 = v(knot)^2 + 2a (A-x(knot))?

Lost.

7. Jan 22, 2014

### ehild

Welcome to PF!

Yes, use the second kinematic equations. Substitute all values given, write XA and xB for the positions. You know that xB=3+xA.

ehild

8. Jan 22, 2014

### grzz

I am going to write your 'v(knot)' as v$_{o}$.

I repeat. Take point A as your zero reference for the distance. Hence the distance at A would have the value of 0.

9. Jan 22, 2014

### grzz

Or follow the suggestion of ehild.

10. Jan 22, 2014

### PHYSStudent098

Thank you.

But does that not still leave me with V^2 = v(knot)^2 + 2(-9.8)(3-A)?

Sorry.

11. Jan 22, 2014

### ehild

You have two equations.

If x =A v =V, given.
If x=B=3+A, v= V/2.

Show these equations.

ehild

Last edited: Jan 22, 2014