vf^2=vi^2 + 2*g*10
solve for vfinal.
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
2 answers
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.
1 answer