# Average force on a baseball when struck by a bat

1. Nov 10, 2014

### brycenrg

1. The problem statement, all variables and given/known data
Baseball(0.145kg) pitch is pitched horizontally at 32m/s, ball pops straight up to a height of 36.5m. If the contact time between bat and ball is 2.5ms, calculate the average force between the ball and the batt during contact.

2. Relevant equations
Favg = Δp/t
KE=KE

3. The attempt at a solution
1/2mv^2 = mgh
v= squar(2gh)
Find Vo when ball leaves bat
((0.145kg*27m/s)-(0.145kg*32m/s))/Tcontact = -290 N

My answer is wrong, in my solution manual it breaks the variables up into x and y. Then it uses Pythagoras equation for the hypotenus. Is that okay to do for forces? i thought that was only used for velocity vectors or.. Ima little confused.

Last edited by a moderator: Nov 10, 2014
2. Nov 10, 2014

### BvU

Force is a vector, momentum is a vector, so $\displaystyle \vec F = {\Delta \vec p \over \Delta t}$ is the equation to work out. It's also the equation you mention under 2. (KE=KE is not all that interesting; why do you list it there ?).

Under 3) attempt at solution you don't continue with the relevant equation, but drag in new stuff that may or may not be applicable. (in fact it is: you need it to find the vertical speed after the hit). But the 27m/s points straight up, whereas the 32 m/s is pointing in the horizontal direction. To go from (32, 0) to (0, 27) you need to change by (-32,+27). The magnitude is found with Pythagoras.

3. Nov 10, 2014

### brycenrg

Oh okay thank you BvU. Whoops i Meant to write P=P or ke+u=ke+u to find v in the direction up. thank you.