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#