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This is the question:

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

Horizontal:

v = 4.0m/s

Vertical:

a = 9.8m/s

All doing with projectile motion

The answer I got was 107.838 meters.

For the baseball it is:

H = ((900sin

H = 29.28 m

V

V

V

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

D

D

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

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/s

^{2}## 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 = ((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/sT = (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 + 19.55D

_{t}= 107.838 mTherefore the total distance between them at the start was about 108 m.

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