How far will the girl run before catching the ball?

[YMMMA]

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?

 

[Chandra Prayaga]

So, its horizontal displacement is x=5(1)= 5m.

Why is that?

 

[YMMMA]

Why is that?

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

 

[gneill]

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?

 

[YMMMA]

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?

 

[YMMMA]

It’s upward

 

[Chandra Prayaga]

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

 

[gneill]

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?

 

[YMMMA]

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

 

[Chandra Prayaga]

Good for you!