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Anna11 [10]
3 years ago
11

A softball player moving 3.89 m/s

Physics
1 answer:
guapka [62]3 years ago
7 0

Answer:

0.119 s is the correct answer to this question.

Explanation:

As mentioned in the question

U=3.89\ m/s

a=-1.44\ m/s^2

S=4.8\ m

Consider the final speed of the softball covering the distance of 4.8m is v

Now using the equation

v^2=U^2+2aS

Putting the value of U, A, S in the previous equation we get

V^2=3.89^2-2\times 1.44\times 4.8\\V=3.7\ m/s\\

Now again using the equation

v=U+at

where U=intial velocity, t= time

Substituting the value of v, U, a

3.7=3.89-1.44\times t\\t=0.119\ s\\

Hence the slide time is 0.119 s

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