Question

sarah is base jumping. if she launches herself from a 50 meter cliff... how fast Will she be moving when she hits the ground assuming her shoot does not open?

Answer 1

Sarah's final velocity is 31.3 m/s downward.

Explanation:

Sarah is in a free-fall motion, which means that there is only one force acting on her: the force of gravity, pusing her downward with a constant acceleration of

$g=9.8 m/s^2$

known as acceleration of gravity.

For a body in free fall, we can apply the suvat equations. In this case, we can use the following equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the vertical distance covered

For this situation, we have

u = 0 is Sarah's initial velocity

$a=9.8 m/s^2$

s = 50 m is the distance covered (the height of the cliff)

Solving for v, we find Sarah's final velocity:

$v=\sqrt{u^2+2as}=\sqrt{0+2(9.8)(50)}=31.3 m/s$

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