What I have got...A pitched ball is hit by a batter at a 45 degree angle and just clears the outfield fence, 98 m away. Assume that the fence is at the same height as the pitch and find the velocity of the ball when it left the bat. Neglect air resistance.
98 m = Velocityinitial * Cos(45) * ( (2 * Velocityinitial * Sin(45) ) / 9.8)
I'm not sure if this is the correct way to solve for Vi, or how to solve it if it is.
I guess I assume no acceleration changes during the flight.
I'm also confused about how I can get the initial velocity with out knowing how long the total flight took. It seems like the there should be direct/indirect? relationship between flight time and initial velocity, but I guess not for some odd reason.
Thanks ahead for any and all help.
Or should it be...
sqrt((98*9.8)/sin(2*45)) just saw some equation like that online, so thought I would post another version.
Ok, I used the equation above and got 30.99032107, and that works when I plug it into the equation I had above. So if it works in that its probably correct. I don't know why, but I have an answer.