Asked by kevin

In the problem, please assume the free-fall acceleration g = 9.80 m/s2 unless a more precise value is given in the problem statement. Ignore air resistance.

A stone is thrown vertically downward from the roof of a building. It passes a window 15.0 m below the roof with a speed of 24.3 m/s. It lands on the ground 4.75 s after it was thrown.

(a) What was the initial velocity of the stone?

(b) How tall is the building?

Answers

Answered by Steve
If building has height H, and stone has initial velocity V,

h(t) = H - 15t - 4.9t^2
h(4.75) = 0 = H - 15(4.75) - 4.9*4.75^2
= H - 181.81

so, H = 181.81m

H-15 = H-15t-4.9t^2
t = .794
v(t) = V - 9.8t
-24.3 = V - 9.8*0.794
V = -16.52m/s
Answered by kevin
thanks for the assistance, but apparently none of the answers are correct according to my homework checker.
Answered by Elena
h=(v²-vₒ²)/2•g,
vₒ= sqrt(v²-2•gvh) =
sqrt(24.3²-2•9.8•15) = 17.22 m/s.
H=vₒ•t+g•t²/2 =
17.22•4.75+9.8•4.75²/2 =192.35 m
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