# Homework Help: Ball thrown up, find max height - Given one velocity in midflight

1. Jan 15, 2014

### ichivictus

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

A ball is thrown upwards from a point near the sidewalk next to a tall office building. At some later time an occupant of the building, exactly 5.1 meters above the point from which the ball was released, sees it pass her window with a speed of 2.9 m/s.

Calculate the maximum height reached by the ball.

You may neglect air resistance. Take the acceleration due to gravity to be 9.82 m/s^2

2. Relevant equations
Kinematic Equations

y = Vot + (1/2)at2
Vf = Vo + at
Vf2 = Vo2 + 2ay
y = t * [ (Vo + Vf) / 2 ]

3. The attempt at a solution

Using Kinematic equation Vf^2=Vo^2+2ay I solve for Vo
(2.9)^2=Vo^2 - 19.64*5.1
Vo = 10.01 m/s

To find max height I set Vf to 0 and solve for y using the same acceleration and the newly found Vo.

0 = 10.01^2 - 19.64y
y=5.1 m

Obviously this doesn't work since 5.1m is where the person at the window sees the ball traveling with velocity. Any help would be great, I tried using the other kinematic equations but it got too messy too quickly.

2. Jan 15, 2014

### nasu

Your value for vo does not look right. I think you forgot to add the (2.9)^2 term.

3. Jan 15, 2014

### ichivictus

That must have been it. I just redid the problem and received an answer a little greater than 12 meters. Not sure where I went wrong exactly, but no worries.

Thanks :)

4. Jan 16, 2014

### PeroK

Here's how I would look at it. Start from the window.

The ball is travelling upwards at 2.9 m/s, so how high does it go from there?

5. Jan 16, 2014

### D H

Staff Emeritus
I suggest you redo the problem again. You didn't show your work, so there's no knowing where you made a mistake.

Hint: Temporarily ignore the fact that the observer was 5.1 meters above the ground. Suppose you threw a ball upwards at 2.9 m/s. How high will it fly given g=9.82 m/s2?