Question #d7002

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Answer 1 · 2015-08-04T21:15:45

I found: c. $14 \frac{m}{s}$ $14 m/s$

Have a look:
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Answer 2 · 2015-08-04T21:31:32

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

$v^{2} = u^{2} + 2 a s$ $v^2=u^2+2as$

In this case the acceleration vector at an angle of $60^{0}$ $60^0$ will be equal to $g cos 60$ $gcos60$

So:

$v^{2} = 0 + 2 \times 9.8 cos 60 \times 20$ $v^2=0+2xx9.8cos60xx20$

$v^{2} = 40 \times 9.8 \times \frac{1}{2}$ $v^2=40xx9.8xx1/2$

$v^{2} = 196$ $v^2=196$

$v = 14 m/s$ $v=14"m/s"$

The mass is irrelevant to the question.