Physics
posted by Laureana .
From a helicopter rising vertically with a velocity of 9 m/s a weight is dropped and reaches the ground in 18 seconds. how high above the ground was the helicopter when the the weight dropped striked the ground.

Upward motion of the weight
h =v₀tgt²/2
v= v₀ gt
v=0 => v₀ =gt => t = v₀ /g =9/9.8 =0.92 s
h = v₀²/2g =9²/2•9.8 =4.13 m
Let h₀ is the height of helicopter when the weight began its motion. Then
for downward motion of the weight
h₀+h =gt₀²/2
t₀ =sqrt{2(h₀+h)/g}
====
t₁=2t+ t₀ = 2v₀/g + sqrt{2(h₀+h)/g} =18 s
sqrt{2(h₀+h)/g} =18  2v₀/g
Square this equation, substitute the given data and calculated height h, and solve for h₀
h₀ = 1280 m
During the motion of the weight, the helicopter moved upwards by the distance
h₁=v₀t₁ = 9•18= 162 m
Therefore,
H = h₀ + h₁ =1280+ 162 = 1442 m . 
h1 = Vo*t + 0.5g*t^2
h1 = 9*18 + 4.9*18^2 = 1426 m Above
ground when bag was released.
h2 = 9*18 = 162 m.. Above point of release.
h = 1426 + 162 = 1588 m. Above ground.