A stone is thrown vertically upward.

[PHYSStudent098]

Homework Statement

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.

Homework 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

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.

 

[grzz]

... Because I'm not given time in the problem statement, I believe I start with Kinematic eq two.

You have chosen the correct formula.

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

 

[PHYSStudent098]

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

 

[grzz]

You take point A as the zero reference for distance.

 

[grzz]

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

 

[PHYSStudent098]

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

Lost.

 

[ehild]

Homework Statement

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.

Homework 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

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.

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

 

[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.

 

[grzz]

Or follow the suggestion of ehild.

 

[PHYSStudent098]

Thank you.

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

Sorry.

 

[ehild]

You have two equations.

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

Show these equations.



ehild