Question

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

Answer 1

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.

Answer: b) 15m below the top of the building