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Art [367]
3 years ago
7

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
Physics
1 answer:
Alik [6]3 years ago
8 0
We use a fundamental kinematic equation as follows:

V = Vo + g*t. 
<span>Tr = (V-Vo)/g = (0-10)/-10 = 1 s. = </span><span>time to reach max. height </span>

<span>Tf = Tr = 1 s. = Fall time or time to fall back to edge of bldg. </span>

<span>3-Tr-Tf = 3-1-1 = 1 s. Below edge of bldg. </span>

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