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barthayn
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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?
Horizontal:
v = 4.0m/s
Vertical:
a = 9.8m/s2
All doing with projectile motion
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.
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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