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](https://tex.z-dn.net/?f=g%3D9.8%20m%2Fs%5E2)
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](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
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](https://tex.z-dn.net/?f=a%3D9.8%20m%2Fs%5E2)
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](https://tex.z-dn.net/?f=v%3D%5Csqrt%7Bu%5E2%2B2as%7D%3D%5Csqrt%7B0%2B2%289.8%29%2850%29%7D%3D31.3%20m%2Fs)
Learn more about free fall:
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