2d Kinematics

  • Thread starter Kildars
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  • #1
Kildars
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A baseball leaves a pitcher's hand horizontally at a speed of 165 km/h. The distance to the batter is 18.3 m. Neglect air resistance.

(a) How long does it take for the ball to travel the first half of that distance?
s
(b) How long does it take for the ball to travel the second half of that distance?
s
(c) How far does the ball fall under gravity during the first half?
m
(d) How far does the ball fall under gravity during the second half?
m

For A I know D and V so I just did v = d/t which translates to t = d/v

It doesn't work.. What am I doing wrong?
 
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Answers and Replies

  • #2
radou
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I suggest you write down the equations for projectile motion. It might be useful to type 'baseball' or 'projectile motion' into the search box, since this is a typical kinematics problem which appears very often here. :smile: Anyway, present some of your work, and we'll be glad to help.
 
  • #3
Kildars
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radou said:
I suggest you write down the equations for projectile motion. It might be useful to type 'baseball' or 'projectile motion' into the search box, since this is a typical kinematics problem which appears very often here. :smile: Anyway, present some of your work, and we'll be glad to help.

:) I did t = d/v which is

Well ask for the time for the first half of the distance of the thrown ball..

So 18.3/2 = 9.15

then t = d/v
t = 9.15/165
Which = .05545 and that is wrong.
 
  • #4
Kildars
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Oh I think I have to convert from km/h to m/s right?
 
  • #5
radou
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Kildars said:
Oh I think I have to convert from km/h to m/s right?

Yes, of course. Watch out for the units.
 
  • #6
Kildars
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Well I did it

165km/h = 45.833 m/s

so t = 9.15/45.833..Which gets you .1996 and that is still wrong.
 
  • #7
Kildars
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radou said:
You got the equation wrong; v = s / t => t = s / v.

t = s/v What's S?
 
  • #8
radou
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Kildars said:
t = s/v What's S?

Oooops, I deleted my last post, because I wrote something stupid, please ignore it. I'm trying to figure this out right now.
 
  • #9
Kildars
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Oh, alright.
 
  • #10
Kildars
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So anyone?
 
  • #11
OlderDan
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Kildars said:
Well I did it

165km/h = 45.833 m/s

so t = 9.15/45.833..Which gets you .1996 and that is still wrong.
Except for the lack of units, this is not wrong. Did you state the problem correctly?
 
  • #12
Kildars
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Nvm, that is right. I was just being stupid and typing it in the wrong box. So a and b are solved, I'm going to try C now.
 
  • #13
Kildars
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So

A = .1996
B = .1996
C = .1952
D = ?

I tried doubling my time and just doing .5(-9.8)(.3992^2)

And got -.78086 Which is incorrect.
 
  • #14
radou
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Kildars said:
So

A = .1996
B = .1996
C = .1952
D = ?

I tried doubling my time and just doing .5(-9.8)(.3992^2)

And got -.78086 Which is incorrect.

This would be correct if the question was 'How far did the ball fall during the whole time of its flight?'. You just have to subtract the answer in C from this.
 
  • #15
Kildars
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Nice, thanks :).
 
  • #16
Kildars
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Time for class now, I'll post more questions if i need help later.
 

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