Physics, asked by scjorgensen2, 3 months ago

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 ritamriyu123
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 beatboxgamming
0

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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