Question #34e08

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Answer 1 · 2018-01-06T14:03:39

The distance the ball can reach is $approx 5.21 m$

The initial speed of ball is

$u = 2.5 ms^-1$

The ball is decelerating due to friction

$a =-0.6ms^-2$ .

When it comes to rest after distance $s$ , the final velocity will be

$v = 0 ms^-1$

According to the position- velocity relation:

$s = (v^2-u^2)/ (2a)$

$s = (0 - 2.5^2)/ (2(-0.6))$ $(ms^-1)^2/( ms^-2)$

$s= -6.25/ -1.2$ $(m^2cancels^-2)/( mcancels^-2)$

$s= 5.208bar33$ $m$

$therefore$ The distance the ball can reach is $approx 5.21 m$