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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
A. 15m above the top of the building
B. 30 m below the top of the building
C. 15 m below the top of the building
D. 30 m above the building

Respuesta :

MissPhiladelphia
We use a fundamental kinematic equation as follows:

V = Vo + g*t. 
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 = Vo*t + 0.5g*t^2. 
d = 10*1 + 5*1^2 = 15 m. <---- OPTION C