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
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?
3 answers
thanks for the assistance, but apparently none of the answers are correct according to my homework checker.
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
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
2 answers