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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 sub...Asked by Breauna
While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 7.65 m/s. The stone subsequently falls to the ground, which is 12.7 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 = 9.81 m/s2. (This is NOT a suggestion to carry out such an experiment!)
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Answered by
Henry
V = Vo + g*Tr = 0
7.65 - 9.81*Tr = 0
9.81Tr = 7.65
Tr = 0.780 s. = Rise time.
h = ho + Vo*Tr + 0.5g*Tr^2
h = 12.7 + 7.65*0.78 - 4.9*0.78^2 = 15.69 m. Above gnd.
a. V^2 = Vo^2 + 2g*h = 0 + 19.6*15.69 = 307.4
V = 17.53 m/s
b. h = 0.5g*t^2 = 15.69
4.9t^2 = 15.69
t^2 = 3.20
t = 1.79 s. = Fall time(Tf).
Tr+Tf = 0.780 + 1.79 = 2.57 s. In air.
7.65 - 9.81*Tr = 0
9.81Tr = 7.65
Tr = 0.780 s. = Rise time.
h = ho + Vo*Tr + 0.5g*Tr^2
h = 12.7 + 7.65*0.78 - 4.9*0.78^2 = 15.69 m. Above gnd.
a. V^2 = Vo^2 + 2g*h = 0 + 19.6*15.69 = 307.4
V = 17.53 m/s
b. h = 0.5g*t^2 = 15.69
4.9t^2 = 15.69
t^2 = 3.20
t = 1.79 s. = Fall time(Tf).
Tr+Tf = 0.780 + 1.79 = 2.57 s. In air.
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