A 0.145-kg 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.

Answer 1

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.

Answer 2

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 =(