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If there is no air resistance, a quarter dropped from the top of New York’s Empire State Building would reach the ground 9.6 s later. (a) What would its speed be just before it hits the ground? (b) How high is the building?

User Bizimunda
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1 Answer

4 votes

Answer:

a) 94.176 m/s

b) 452.04 m

Step-by-step explanation:

t = Time taken by the quarter to reach the ground = 9.6 s

u = Initial velocity

v = Final velocity

s = Displacement

a)


v=u+at\\\Rightarrow v=0+9.81* 9.6\\\Rightarrow v=94.176 m/s

Speed of the quarter just before it hits the ground is 94.176 m/s

b)


s=ut+(1)/(2)at^2\\\Rightarrow s=0* t+(1)/(2)* 9.81* 9.6^2\\\Rightarrow s=452.04\ m

The building is 452.04 m high

User Maloo
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