While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 6.85 m/s. The stone subsequently falls to the ground, which is 19.1 m below the point where the stone leaves his hand.
At what speed does the stone impact the ground? Ignore air resistance and use =9.81 m/s2 for the acceleration due to gravity.
Answers
Answered by
2
h = ho + -(Vo^2)/2g
h = 17.1 + 6.93^2/19.6 = 19.6 m. Above
gnd.
a. V^2 = Vo^2 + 2g*h = 0 + 19.6*19.6 =
382.2
V = 19.6 m/s.
b. Tr = -Vo/g = -6.93/-9.8 = 0.707 s. =
Rise time.
h = 0.5g*t^2 = 19.6 m.
4.9t^2 = 19.6
t^2 = 4
Tf = 2 s. = Fall time.
Tr+Tf = 0.707 + 2 = 2.707 s = Time in air.
Answered by
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YOUR ANSWER IS
Explanation:
h = ho + -(Vo^2)/2g
h = 17.1 + 6.93^2/19.6 = 19.6 m. Above
gnd.
a. V^2 = Vo^2 + 2g*h = 0 + 19.6*19.6 =
382.2
V = 19.6 m/s.
b. Tr = -Vo/g = -6.93/-9.8 = 0.707 s. =
Rise time.
h = 0.5g*t^2 = 19.6 m.
4.9t^2 = 19.6
t^2 = 4
Tf = 2 s. = Fall time.
Tr+Tf = 0.707 + 2 = 2.707 s = Time in air.
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