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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?

1. The problem statement, all variables and given/known data

Horizontal:

v = 4.0m/s

Vertical:

a = 9.8m/s^{2}

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 = ((900sin^{2}(53))/19.6)

H = 29.28 m

V_{f}^{2}= (30sin(53))^{2}+(2*9.8*29.28)

V_{f}^{2}= 1138.07m^{2}/s^{2}

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}= d_{1}+ d_{2}

D_{t}= 88.279 + 19.55

D_{t}= 107.838 m

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

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# Homework Help: Is this correct answer?

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