Projectile Motion: Calculating Distance Between Batter and Outfielder"

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The discussion centers on calculating the distance between a batter and an outfielder in a projectile motion scenario involving a baseball hit at 30 m/s at a 53-degree angle. The calculations provided yield a total distance of approximately 108 meters between the batter and the outfielder. Key equations used include those for horizontal and vertical motion, with the outfielder's running speed factored into the final distance. A critique is noted regarding the calculation of final velocity, as it should not exceed the initial speed of the baseball. The overall conclusion confirms the initial distance based on the projectile motion analysis.
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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?

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

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.
 
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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.
 
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