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An object is thrown upward from the edge of a tall building with a velocity of 10 m/s. Where will the object be 3 s after it is thrown? Take g=10ms^2

A15m above the top of the building
B30 m below the top of the building
C15 m below the top of the building
D30 m above the building

Verified Answer :

MissPhiladelphia

We use a fundamental kinematic equation as follows:

V = Vo + gt. 
Tr = (V-Vo)/g = (0-10)/-10 = 1 s. = time to reach max. height 

Tf = Tr = 1 s. = Fall time or time to fall back to edge of bldg. 

3-Tr-Tf = 3-1-1 = 1 s. Below edge of bldg. 

d = Vot + 0.5gt^2. 
d = 101 + 5*1^2 = 15 m. <---- OPTION C

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