Baseball and Average Force Question

[kliang1234]

Question:
A 0.165kg baseball pitched horizontally at 34.5 m/s strikes a bat and is popped straight up to a height of 40.0m.

a) If the contact time between bat and ball is 1.0ms, calculate the magnitude of the average force between the ball and bat during contact.

I got this part right. I used Pythagorean theorem, kinematics, and momentum to solve for the average force.
I get 7331N as average force

b) Find the direction of the average force on the ball.

This is where i can't seem to get it right.
I've set up the triangle with velocity vectors vx and vy. Then i used tan^-1 to get an angle. This was incorrect.

Given that vx=34.5m/s and finding that vy=28m/s

i used tan^-1 (Vup/Vpitched) and that angle was incorrect
I also tried tan^-1(Vpitched/Vup) and that angle was also incorrect

any hints on how to find the direction of average force?
The angle the question is looking for is the angle from the initial direction of motion of the ball.

Thanks

 

[Doc Al]

The angle the question is looking for is the angle from the initial direction of motion of the ball.

Draw a diagram showing the change in velocity. (That will give you a triangle like you described.) Then describe the direction of that force with respect to the original velocity of the ball. (That last part is the key.)

 

[kliang1234]

I tried that on my first attempt.
Since its Vf-V0,
i put

Vf
^ --------> V0
|
|
|

I get the angle between V0 and the hypotenuse to be 39°
But since its asking for the angle relative to the intial motion.. would it actually be -39° ?
or 180+39 = 219° ?

Thanks

 

[Doc Al]

I tried that on my first attempt.
Since its Vf-V0,
i put

Vf
^ --------> V0
|
|
|

What you've drawn actually shows Vf + V0. You need Vf - V0. (Reverse the direction of V0.)

I get the angle between V0 and the hypotenuse to be 39°

That part's correct, but you don't have the hypotenuse in the right direction.

 

[kliang1234]

are you sure?
the ball is thrown towards the bat where it then gets popped up vertically.

^Vf
|
|
|
|
<---------V0 the triangle above would be for Vf+Vo

By reversing the Vo and putting its tail to Vf's head, i get
Vf-V0 and that's below----------->V0
^
|
|
|
|Vf

 

[Doc Al]

are you sure?
the ball is thrown towards the bat where it then gets popped up vertically.

^Vf
|
|
|
|
<---------V0


the triangle above would be for Vf+Vo

Does Vo point to the left? So the ball was thrown to the left? (From your earlier diagram, I thought Vo pointed to the right.)

By reversing the Vo and putting its tail to Vf's head, i get
Vf-V0 and that's below


----------->V0
^
|
|
|
|Vf

If the original velocity of the ball was to the left, then that would be correct.