physics
posted by mt .
While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 7.53 m/s. The stone subsequently falls to the ground, which is 14.7 m below the point where the stone leaves your hand. At what speed does the stone impact the ground? How much time is the stone in the air? Ignore air resistance and take g = 9.80 m/s2.

h = ho + (V^2Vo^2)/2g
h = 14.7 + (0(7.53)^2)/19.6 = 17.6 m.
Above gnd.
V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*17.6 = 345
V = 18.6 m/s.
V = Vo + g*t
Tr = (VVo)/g = (07.53)/9.8=0.768 s.=
Rise time.
h = Vo*t + 0.5g*t^2 = 17.6 m.
0 + 4.9t^2 = 17.6
t^2 = 3.59
Tf = 1.89 s. = Fall time.
T = Tr + Tf = 0.768 + 1.89 = 2.66 s. =
Time in air.
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