While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of

Question

While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of

6.63 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 = 9.8 m/s2.

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1 answer
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Answer 1

Anonymous answered
3 years ago

constant a = -g = -9.81 m/s^2

v = Vi + a t
where Vi = 6.63 and a = -9.81
so
v = 6.63 - 9.81 t

h = Hi + Vi t + (1/2) a t^2
where Hi = 12.7
so
h = 12.7 + 6.63 t - 4.9 t^2

Now
at ground h = 0
so
4.9 t^2 - 6.63 t -12.7 = 0
t = -1.06 (before you threw it) and t = 2.42 seconds (use that)
v = 6.63 - 9.81 t = -17.1 m/s