# Finding the initial velocity of a projectile, given time, an angle, and distance

1. Jan 18, 2010

### science_rules

1. The problem statement, all variables and given/known data
A baseball is thrown at 20 degrees to the horizontal. The ball travels a horizontal distance of 18 meters when it returns to the same level as the instant it had been thrown. The total time it traveled between these points is 2.1 seconds.
Determine the initial velocity (Vo) of the ball at its starting point.

2. Relevant equations
Range (distance) equation = (Vox)t, t = total time = 2.1 seconds
Vox = Vocos20

3. The attempt at a solution
(RIGHT WAY?)
R = (Vox)t
18 meters = (Vox)t = (Vocos20)(2.1 seconds) = (Vo)(1.973)
then: 18 meters/1.973 = 9.123 m/s initial velocity

I was initially thinking of solving it by using the time equation to solve for Vo (but then i realized i needed a different equation to incorporate the distance information given):
t = Vfy - Voy / -9.8 m/s^2 = 2.1 seconds (WRONG WAY) there was no way i could find Vo using this way since there was no way to incorporate the distance info given into the time equation, so i used the Range equation instead. I assume there is no need to find Vfy specifically for this problem.

Question: Would Vfy (being at the same level as Voy was when the ball started) be the same number as Voy, only a negative number (because of downward velocity)?? Since the ball's velocity decreases during the first half of the trajectory, then increases during the second half, it seems to me that it would approach the exact same velocity at a given point on the other side of its trajectory.

2. Jan 18, 2010

### Yitzach

That is correct, assuming no air resistance.