An object is thrown upward from the edge of a tall building with a velocity of 10m/s. Where will the object be 3secs after it is thrown? Take g = 10m/s^2
a) 30m above the building
b) 15m below the top of the building
c) 15m above the top of the building
d) 30m below the top of the building
1
Expert's answer
2017-05-27T04:37:10-0400
The low of motion of and object due to gravity is y(t)=v_0 t-(gt^2)/2=10t-5t^2. Y axis is directed upwards. The origin is placed on the top of the building. Coordinate of the object after 3secs it is thrown: y(3)=-15. Thus, the object will be 15m below the top of the building.
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