Is this correct answer?

  • Thread starter barthayn
  • Start date
  • #1
87
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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.
 
Last edited:

Answers and Replies

  • #2
2,014
289
Yes, it's Ok. Your calculation of [itex] V_f [/itex] seems to be wrong, The ball will never go faster then 30 m/s. You're also not using it.
 

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