While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 6.71 m/s. the stone subsequently falls to the ground, which is 18.9 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

A) The speed at which the stone impacts the ground is; v = 20.39 m/s

B) The time at which the stone is in the air is; t = 1.39 s

We are given;

Initial velocity; u = 6.71 m/s

Distance; s = 18.9 m

Acceleration due to gravity; g = 9.81 m/s²

    To calculate the speed at which the stone impacts the ground, we will use newtons' third equation of motion which is;

v² = u² + 2gs

Where;

v = final velocity

u = initial velocity

a = acceleration due to gravity

s = distance

A) Plugging in the relevant values gives;

v² = 6.71² + 2(9.81 × 18.9)

v² = 415.8421

v = √415.8421

v = 20.39 m/s  

2) We can find the time at which the phone is in the air using newton's first equation of motion which is; v = u + gt

Plugging in the relevant values gives;  

 20.39 = 6.71 + 9.81t

9.8t = 20.39 - 6.71

9.81t = 13.68

t = 13.68/9.81

t = 1.39s

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are there options? cause you ain’t. real specific

no

sodium is solid at rt