# Finding time a ball is in the air

1. May 13, 2009

### mcryder16

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

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 15.44 m/s when the hand is 1.61 m above the ground. How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)

2. Relevant equations

I need help starting this problem

3. The attempt at a solution

2. May 13, 2009

### LowlyPion

See:
https://www.physicsforums.com/showpost.php?p=905663&postcount=2

3. May 13, 2009

### fallen186

$$v_{f}=v_{o}+at$$
$$0m/s = 15.44 m/s +(-9.81m/s^2)*t$$
$$t = 1.574 s$$

$$v_{f}^2=v_{o}^2+2a*x$$
$$0 = 15.44^2m/s + 2 *(-9.81m/s^2) * x$$
$$x = 12.15 m$$
This is how high the ball travels from the hand.

-------------When the ball is returning to earth-----------
$$x = x_{0} +v_{0}t+\frac{1}{2}at^{2}$$
$$x = 12.15m + 1.61m = 13.76m$$
$$13.76m = 0m + 0m/s*t + \frac{1}{2} * 9.81m/s^2*t^2$$
$$27.52 = 9.81m/s^2*t^2$$
$$2.805 = t^2$$

$$t = 1.675 s$$

$$t_{total} = t_{ascent} +t_{decend}$$

$$t_{total} = 1.975s + 1.574 s$$

$$t_{total} = 3.549s$$