Question

A softball player moving 3.89 m/s

Answer 1

Answer:

0.119 s

Explanation:

Given that

$U=3.89\ m/s\\a=-1.44\ m/s^2\\S=4.8\ m$

Lets take final speed of the softball after covering 4.8 m = V

We know that

$V^2=U^2+2aS\\V^2=3.89^2-2\times 1.44\times 4.8\\V=3.7\ m/s$

Also We know that

$V=U+at$

Putting the value of V ,U and a in the previous equation  We get

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

Therefore slide time will be 0.119 s