# How far will the girl run before catching the ball?

Below

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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Chandra Prayaga
So, its horizontal displacement is x=5(1)= 5m.
Why is that?

Why is that?
Is it not 0 fotr the final horizontal velocity? Should I consider the motion of the girl?

gneill
Mentor
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?

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?

Chandra Prayaga
What is the direction of the inotial velocity of the ball?

It’s upward

Chandra Prayaga
So why are you using that to calculate the horizontal displacement?

gneill
Mentor
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?

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

• gneill
Chandra Prayaga
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