Question #98a41

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Answer 1 · 2016-10-07T01:09:56

Let

$u-> "Initial velocity of the ball"=2.2" m/s"$

$v-> "Final velocity of the ball"=0" m/s"$

$s-> "Length of ascent along the slope"=3.2m$

$t-> "Time of ascent along the slope"$

Since the motion occurs with constant retardation,we can write

$1/2(u+v)xxt=s$

$=>1/2(2.2+0)xxt=3.2$

$=>t=3.2/1.1~~2.9s$