Answer :
The height of the building is 145.5 m
Calculation
It is given that
[tex]y_{last}[/tex]= h/3
[tex]t_{last}[/tex]= 1s
g= 9.8 m[tex]s^{-2}[/tex]
In first stage,
t1= (t-1)s y1 = 2h/3
vi1 = 0
according to the kinematic equation,
y1= v1t1 + 1/2 g[tex]t1^{2}[/tex]
then substituting all the values, we have
2h/3 = 1/2(9.8)[tex](t-1)^{2}[/tex]
⇒ h = 7.35[tex](t-1)^{2}[/tex] → 1
Then the final velocity can be calculated by
vf1=[tex]vi^{2}[/tex]+ g(t - 1)
vf1= 9.8(t - 1)
In the second stage,
t2= 1s
y2= h/3
g=9.5 m[tex]s^{-2}[/tex]
vi2 = vf1 = 9.8(t - 1)
Acoording to kinematic equation,
y2 = vi2 t2 + 0.5g[tex]t1^{2}[/tex]
substituting the values,
h/3 = 9.8(t - 1) + 4.9
⇒ h= 29.4(t - 1) + 14.7 → 2
From equation 1 and 2, we get,
7.35[tex](t-1)^{2}[/tex] = 29.4(t - 1) + 14.7
if we assume (t-1) = R, we get
7.35[tex]R^{2}[/tex] - 29.4R - 14.7 =0
⇒R = 4.45
t = R+1 = 5.45s
substituting the value of t in equation 1, we have
h= 7.35[tex](4.45)^{2}[/tex] = 145.5m
Thus, the height of the building is 145.5m.
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