Projectile Motion: Calculating Distance Between Batter and Outfielder"

[barthayn]

This is the question:

A baseball is hit at 30m/s at an angle of 53.0^o 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?

Homework Statement

Horizontal:
v = 4.0m/s

Vertical:
a = 9.8m/s²

Homework Equations

All doing with projectile motion

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

V_f² = (30sin(53))²+(2*9.8*29.28)
V_f² = 1138.07m²/s²
V_f = 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

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

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

 

[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.