In a relief operation mission involving a helicopter h distance above the ground traveling with a horizontal velocity vx. What should be the expression for the ho distance dy at which you release the relief package so that it will arrive to the survivors at the right place? (Neglect the effect of air resistance) Solution To determine this, we must first derive the time it takes for the relief package to reach the survivors. We use this equation -h = Vinitial-yt + (1/2)ay If we just drop the package from the helicopter, the equation above becomes + (1/2)ay Substituting ay = -g then simplifying results to t = sqrt( which is the time it takes for the object to reach the ground. Since the package will just travel at a constant velocity in the x-axis, thus dy = Vxt Substituting the time taken by the package to reach to ground results to: dy = ( )( sqrt ( ))
In a relief operation mission involving a helicopter h distance above the ground traveling with a horizontal velocity vx. What should be the expression for the ho distance dy at which you release the relief package so that it will arrive to the survivors at the right place? (Neglect the effect of air resistance) Solution To determine this, we must first derive the time it takes for the relief package to reach the survivors. We use this equation -h = Vinitial-yt + (1/2)ay If we just drop the package from the helicopter, the equation above becomes + (1/2)ay Substituting ay = -g then simplifying results to t = sqrt( which is the time it takes for the object to reach the ground. Since the package will just travel at a constant velocity in the x-axis, thus dy = Vxt Substituting the time taken by the package to reach to ground results to: dy = ( )( sqrt ( ))
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