The final velocity of the candy is 57.7 m/s
Explanation:
The motion of the candy is a free fall motion, since it is subjected only to the force of gravity, so it is a uniformly accelerated motion and therefore we can use the following suvat equation:
[tex]v^2-u^2=2as[/tex]
where
v is the final velocity
u is the initial velocity
[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity
s is the vertical displacement
In this problem, we have:
u = 0 (the candy is dropped from rest)
s = 170 m (the vertical displacement is the height of thr monument)
Solving for v, we find the velocity of the candy as it hits the ground:
[tex]v=\sqrt{u^2+2as}=\sqrt{0+2(9.8)(170)}=57.7 m/s[/tex]
Learn more about free fall motion:
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