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How far will the girl run before catching the ball?

  • Thread starter YMMMA
  • Start date
  • #1
156
10

Homework Statement


Below

Homework Equations


Y=1/2 *gt^2
X=Vt
Vf=Vi+gt

The Attempt at a Solution


The ball will have the same final velocity as Its initial. So, using the last equation 0=5+gt, where g is 10. Therefore, time to reach the maximum height is half seconds. To reach the same height it started from, kt will take another half a second, Totaltime is 1 second. So, its horizontal displacement is x=5(1)= 5m. Since the same time applies for the motion of the girl, she will have moved a distance of 3m. Is my reasoning correct?
 

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Answers and Replies

  • #2
Chandra Prayaga
Science Advisor
649
148
So, its horizontal displacement is x=5(1)= 5m.
Why is that?
 
  • #3
156
10
Why is that?
Is it not 0 fotr the final horizontal velocity? Should I consider the motion of the girl?
 
  • #4
gneill
Mentor
20,792
2,770
Is it not 0 fotr the final horizontal velocity?
No. It's constant.
Should I consider the motion of the girl?
What precisely does the question ask you to find?
 
  • #5
156
10
No. It's constant.

What precisely does the question ask you to find?
How far will she run before catching it?

It is constant for the horizontal direction and zero vertically. Sorry, I confused my self. Here, the horizontal distance is the initial velocity times the time. What’s incorrect?
 
  • #6
Chandra Prayaga
Science Advisor
649
148
What is the direction of the inotial velocity of the ball?
 
  • #7
156
10
It’s upward
 
  • #8
Chandra Prayaga
Science Advisor
649
148
So why are you using that to calculate the horizontal displacement?
 
  • #9
gneill
Mentor
20,792
2,770
So, its horizontal displacement is x=5(1)= 5m
The above statement is incorrect. The ball does not have a horizontal velocity component of 5 m/s. Your answer for the girl's horizzontal displacement is correct. Surely they need to have the same horizontal displacement if she's to catch the ball?
 
  • #10
156
10
Aha, now I know I messed up. Yes they both must have the same horizontal displacement and I used the vertical velocity instead . Got it, Thank you all
 
  • #11
Chandra Prayaga
Science Advisor
649
148
Good for you!
 

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