Answer:
a) 94.176 m/s
b) 452.04 m
Explanation:
t = Time taken by the quarter to reach the ground = 9.6 s
u = Initial velocity
v = Final velocity
s = Displacement
a)
[tex]v=u+at\\\Rightarrow v=0+9.81\times 9.6\\\Rightarrow v=94.176 m/s[/tex]
Speed of the quarter just before it hits the ground is 94.176 m/s
b)
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 9.81\times 9.6^2\\\Rightarrow s=452.04\ m[/tex]
The building is 452.04 m high
