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Sagot :
The rock's maximum speed just as it hits the ground is 28m/s.
Given that the initial speed of the rock is u = -14m/s. (speed taken as negative because the rock thrown down from the cliff)
The height of the tower is h=30m.
Taking height of the tower as origin, then final position of the rock is s= -h= -30m.
On applying the second equation of motion:
s= ut +¹/₂ at²
-30 = -14t+0.5×(-9.8)t²
4.9t²+14t-30=0
On Solving the equation we get, t=1.43s.
To calculate that final speed when rock hits the ground, we apply first equation of motion:
v=u+gt
v=-14+(-9.8)(1.43)
v=-28m/s.
To learn more about Equations of motion, refer this link
https://brainly.com/question/5955789
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