# Homework Help: Is this correct answer?

1. Feb 10, 2010

### barthayn

This is the question:

A baseball is hit at 30m/s at an angle of 53.0o with the horizontal. Immediately, an outfielder runs at 4.00m/s toward the infield and catches the ball at the same height it was hit. What was the original distance between the batter and the outfielder?

1. The problem statement, all variables and given/known data
Horizontal:
v = 4.0m/s

Vertical:
a = 9.8m/s2

2. Relevant equations
All doing with projectile motion

3. The attempt at a solution
The answer I got was 107.838 meters.

For the baseball it is:

H = ((900sin2(53))/19.6)
H = 29.28 m

Vf2 = (30sin(53))2+(2*9.8*29.28)
Vf2 = 1138.07m2/s2
Vf = 33.88m/s

T = (60sin53)/9.8
T = 4.88s

d = 30cos53*4.88
d = 88.279m

For the running it is:
d = 4*4.88
d = 19.55 m

Dt = d1 + d2
Dt = 88.279 + 19.55
Dt = 107.838 m

Therefore the total distance between them at the start was about 108 m.

Last edited: Feb 10, 2010
2. Feb 10, 2010

### willem2

Yes, it's Ok. Your calculation of $V_f$ seems to be wrong, The ball will never go faster then 30 m/s. You're also not using it.