I'm trying to solve that equation right now... kinda of confusing.
But back to the original post - ok so if the ball reached it's peak at a horizontal distance of 12.8m, then if the trajectory of the ball is symetrical, the ball traveled 25.6m when it hit the ground - correct?
Yes horizontal velocity. So the ball reached its horizontal peak at 12.8m? I thought it was 6.4 because I had originally thought that 12.8m was the total distance the ball had traveled untill it hit the ground. -9.8m/s^2 in the equation had already gave me the vertical peak correct?
A ball is thrown into the air with an initial vertical velocity of 5.0m/s in addition to its horizontal velocity of 25.0m/s, how far would the ball travel horizontally before it reached it's peak?
vf = vi + at
I've also tried vf = 1/2g +at...