Question #d7002

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Answer 1 · 2015-08-04T21:15:45

I found: c. #14 m/s#

Have a look:
enter image source here

Answer 2 · 2015-08-04T21:31:32

(c). #14"m/s"#

#v^2=u^2+2as#

In this case the acceleration vector at an angle of #60^0# will be equal to #gcos60#

So:

#v^2=0+2xx9.8cos60xx20#

#v^2=40xx9.8xx1/2#

#v^2=196#

#v=14"m/s"#

The mass is irrelevant to the question.