# Homework Help: Throwing a ball into the air

1. Sep 10, 2009

### Amel

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

The problem: A woman standing on a diving board 21 meters above a pool tosses a ball upwards at a speed of 19 meters/second. The ball goes up and then down, splashing into the water below. How long was the ball in the air? What is the ball’s final velocity?

2. Relevant equations

v2= v02+2a(x-x0)
X = X0 +V0t+(1/2)at2

3. The attempt at a solution

So I figured out the height of the ball by using v2= v02+2a(x-x0) and having v = 0 because thats what it is at the peak and re aranged it so it gives me X. I got 39.25 meters. But I don't know how to get the time the ball is in the air. I assume I have to possibly use X = X0 +V0t+(1/2)at2 but im not sure how to get time out of it.

2. Sep 10, 2009

### 206PiruBlood

You are correct in that your position function allows you to solve for time. If you take the pool to be at height zero meters, then what is your initial height? You know the acceleration due to gravity and the initial velocity. Put these figures into your equation of motion and you are left with a function of the variable t.

3. Sep 10, 2009

### Amel

Ok so I did t=V0+/- The square root of ((V02 + 2y0g))/g and got 4.94 seconds, The equation came form a problem in the book with a similiar problem. But it kind of seems to short of a time. I feel like its not counting the time the ball is travelling up or something.

Anyway is that right? want to know before I do the last part of the equation.

4. Sep 10, 2009

### 206PiruBlood

Use the equation $$x(t)= x_0{}+v_0t-\stackrel{1}{2}gt^{2}$$.

My bad, misinterpreted your comment. I just put it in a calculator and got 4.775. Did you use 19m/s as your initial velocity and -9.8 for acceleration? It is much easier to find the time first then the final velocity second.

Last edited: Sep 10, 2009