Asked by Anonymus
A ball is thrown up from a 10 -m-high cliff with an initial speed of 2.00 m/s. What is the speed of the ball just before it hits the ground? (The ground is defined as the bottom of the cliff.) Neglect air resistance
Answers
Answered by
bobpursley
vf^2=vi^2 + 2*g*10
solve for vfinal.
solve for vfinal.
Answered by
henry2,
V^2 = Vo^2 + 2g*h = 0.
4 + (-19.6)h = 0,
h = 0.20 m. above cliff.
ho + h = 10 + 0.2 = 10.2 m. above gnd.
V^2 = Vo^2 + 2g*(ho+h) = 0 + 19.6*10.2 = 200,
V = 14.14 m/s.
4 + (-19.6)h = 0,
h = 0.20 m. above cliff.
ho + h = 10 + 0.2 = 10.2 m. above gnd.
V^2 = Vo^2 + 2g*(ho+h) = 0 + 19.6*10.2 = 200,
V = 14.14 m/s.
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