Posted by **Laureana** on Sunday, August 4, 2013 at 10:44am.

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.

- Physics -
**Elena**, Sunday, August 4, 2013 at 5:55pm
Upward motion of the weight

h =v₀t-gt²/2

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 .

- Physics -
**Henry**, Sunday, August 4, 2013 at 7:48pm
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.

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