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 =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 .
so what is the right solution?? is it by elena or henry??
To determine the height above the ground at which the weight was dropped, we can use the equation of motion for an object in free fall:
h = (1/2)gt^2
where:
h is the height
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time
In this case, the weight took 18 seconds to reach the ground. Plugging in the values into the equation, we get:
h = (1/2)(9.8 m/s^2)(18 s)^2
Calculating this expression gives us:
h ≈ (1/2)(9.8 m/s^2)(324 s^2)
h ≈ 1584 m
Therefore, the weight was dropped from a height of approximately 1584 meters above the ground.
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.