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.