Physics Problem?

Question

A baseball is hit at 30m/s at an angle of 53 degrees with the horizontal. Immediately an outfielder runs 4 m/s toward the infield and catches the ball at the same height it was ht.

What was the original distance between the batter and the outfielder?

Answer 1

Here are the equations of motion for the ball

vy(t)=30*sin(53)-g*t
x(t)=30*cos(53)*t

the ball will reach the same height it was hit in twice the time to apogee
at apogee, vy(t)=0
t=30*sin(53)/9.81
so the ball travels
x(range)=2*30^2*sin(53)*cos(53)/9.81
the ball travels 88 meters
In the same time, the outfielder travels
4*t=4*2*30*sin(53)/9.81
=19.5 m

The original distance was
107.5 m

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