Answer:
Height of the building = h = 30 m
Initial speed, when dropping = 0
Speed just before hitting the ground = v =?
Acceleration due to gravity = g =9.8 ms-2 or if you wish,you can use 10 ms-2
Solution
Applying law of conservation of energy:
Kinetic energy just before hitting the ground = the potential energy when releasing it.
1/2mv^2 = mgh
v =sqrt of (2gh) = sqr(2x9.8x30) =24.24 =24 m/s
(Answer corrected to two significant figures, because the given data is correct to 2 significant figures. )
Alternatively you can use,
v^2 = u^2 + 2as
Where v = ?, u =0, a =g = 9.8 or 10 ms^2
Substitute the values and get the answer.
