Can a Field Player Catch a Baseball Hit at 27 m/s and 32 Degrees?

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SUMMARY

A baseball is hit at a velocity of 27 m/s at an angle of 32 degrees, reaching a height of 1 meter. A field player, positioned 50 meters from home plate, must account for a reaction time of 0.5 seconds to catch the ball at the same height from which it was hit. The calculations indicate that the player needs to run at a velocity of 7.57 m/s to successfully intercept the ball, factoring in the time it takes for the ball to reach him, which is approximately 2.99 seconds.

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Homework Statement


A baseball is hit at a height of 1m with an initial velocity of 27 m/s at 32 degrees above the horizontal. A field player is located 50m from the home plate along the line of flight.

(a) If he runs, is there a chance he could catch it? what's the velocity?
Assume his reaction time is 0.5s, and that he would catch it at the initial level. MAIN ISSUE: (i don't know what he means by "initial level". initial level of the pitcher, or field player?)

Homework Equations


kinematics equations:
Xf = Xi + ViT + 1/2at^2
Vf^2 = Vi^2 + 2a(delta x)
Vf = Vi + a(delta T)

The Attempt at a Solution


0 = 1 + 14.308T - 4.8T^2
...
T = 2.99s
66.86 = 50 + Vi(2.99s)
...
basically, I got Vi as 7.57m/s instead of 7... maybe it has to do with the "initial position" that i didn't get?
 
Last edited:
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Initial level is simply the height from the ground at which the baseball is hit. Don't forget to take into account the fielder's reaction time.
 

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