While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 5.55 m/s. the stone subsequently falls to the ground, which is14.7 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.

what is the impact speed: /s

what is the elapsed time:

Let's solve first for the impact speed. The impact speed or velocity of any free-falling body follows the formula written below:

v = √2gy
where
y is the height of the free fall

Now, the length of y includes the distance it took for the upward motion, and the free falling motion that covers the maximum height plus the remaining distance of 14.7 ft. Let's compute first the maximum height reached:

Hmax = v₀²/2g, where v₀ is the initial velocity
Hmax = (5.55 m/s)²/2(9.81 m/s²) = 1.57 m
Thus,
y = 1.57 m + 14.7 m = 16.27 m
v = √2(9.81 m/s²)(16.27 m)
v = 17.87 m/s

For the second question, there are two sections of the total time. The first is the time for the upward motion:
t₁ = 2v₀/g = 2(5.55 m/s)/(9.81 m/s²) = 1.13 s
The second section is the time for the free fall:
t₂ = √2y/g = √2(16.27 m)/9.81 m/s² = 1.82 s
Thus, the total time is:
Time = 1.13 s + 1.82 s = 2.95 s

Not what goes around on earth but the universe too.