Question

A softball player moving 3.89 m/s

Answer 1

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