physics for scientists and engeneers I
posted by Johnathon on .
A 0.145kg baseball pitched horizontally at 27 m/s strikes a bat and is popped straight up to a height of 43 m. If the contact time between the bat and the ball is 2.35 ms, calculate the average force [exerted by the bat on the ball] during contact. [Let the positive axis lie along the line from the batter to the pitcher, with the batter at the origin.] Please give magnitude and direction.

assuming the 27 was really horizontal, then the change in velocity has to be determined.
To get 43m high (that is pretty high)...
1/2 m vi^2=mg(43)
vi=sqrt (86*9.8)
change momentum=massball*deltaV
ok,to get deltaV, deltaV=sqrt(86*9.8) j +27i)
then force= changemometum/time
you get direction from the velocity coordinates. 
im so lost that its not even funny... idk what im doing and it wouldve been better if this person wouldve used numbers instead of letters =(