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After three seconds, what is the skydiver's velocity? Show your work and include direction.
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use this equation\[v=v_0+at\]the skydiver jumps out of the plane and falls for 3 seconds so she started from rest in the plane and fell for 3 seconds. so since she started from rest her initial velocity is going to be 0. we know since the person is skydiving they're doing it on earth so their acceleration is just 9.8 m/s^2 in the downwards direction and t is just 3 seconds. so it would just be\[v=(0)=(9.8\frac{ m }{ s^2 })(3s)\]
cripes. messed up that last equation. it's v=0+(9.8)(3)
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