Solution

a) the speed of the object when t = 3 second
Dik:t = 3 s
a = g = 10 m/s²
V_o = 0 m/s
Dit:V_t ?
Vt = Vo + at
Vt = 0 + (10)(3) = 30 m/s

b) the distance of the object during 3 second
S = Vot + 1/2at²
S = (0)(t) + 1/2 (10)(3)²
S = 45 meter

c) height of the current object t = 3 second
height of the current object t = 3 sekon is high initially reduced distance traveled objects.
S = 100 − 30 = 70 meter

d) the speed of the object when they arrive on the ground
Vt² = Vo2 + 2aS
Vt² = (0) + 2 aS
Vt = √(2aS) = √[(2)(10)(100)] = 20√5 m/s