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blondinia [14]
2 years ago
7

A major-league pitcher can throw a baseball in excess of 41.0 m/s. If a ball is thrown horizontally at this speed, how much will

it drop by the time it reaches a catcher who is 17.0 m away from the point of release?
Physics
1 answer:
SSSSS [86.1K]2 years ago
5 0

Answer:

Ball will drop by 0.82 meter.

Explanation:

Horizontal speed = 41 m/s

Horizontal displacement = 17 m

Horizontal acceleration = 0 m/s²

Substituting in s = ut + 0.5at²

    17 = 41 t + 0.5 x 0 x t²

     t = 0.41 s

Now we need to find how much vertical distance ball travels in 0.41 s.

Initial vertical speed = 0 m/s

Time = 0.41s

Vertical acceleration = 9.81 m/s²

Substituting in s = ut + 0.5at²

    s = 0 x 0.41 + 0.5 x 9.81 x 0.41²

    s = 0.82 m

So ball will drop by 0.82 meter.

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Answer:

The distance is D  =  2.6 \ m

Explanation:

From the question we are told that

    The wavelength of the light is  \lambda  =  476.1 \ nm  =  476.1 *10^{-9} \ m

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        D  =  2.6 \ m

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The first part of the question is:

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