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A pilot is attempting to deliver emergency good and first-aid supplies to an isolated northern community that has suffered severe flooding. The plane has a horizontal velocity of 1.40 X 10^2 km/h, as it flies at an altitude of 1.80 X 10^2 metres. The community is situated on a dry patch of land that is about 72m X 72m.
a) If the supplies are released just as the plane flies directly overhead, will they touch down on land or in water? Justify your response with calculations that show exactly where the package will land.
b) When should the supplies be released so that they touch down very close to the centre of the dry patch of land? Answer in terms of distance from the target, not time.
Equations used:
Vertical distance = 1/2(acceleration)(delta t^2)
Horizontal distance = (horizontal velocity)(delta t)
c^2 = a^2 + b^2
a) I know that the vertical displacement is given, but the horizontal displacement is not. In order to find the horizontal component displacement, I would need to find the change in time time. I take into account that there is no initial velocity, so I can use the vertical distance = 1/2(acceleration)(delta t^2) rather than Vertical distance =(initial velocity)(delta t)+1/2(acceleration)(delta t^2) equation. Since I'm given vertical distance = 180m, I can isolate and solve for time which is approximately 6.1 seconds. Now I can find the horizontal distance that is traveled:
Horizontal distance = (horizontal velocity)(change in time) = (38.889 m/s after conversion)(6.1) = 235.6 m forward. Now that I have both the horizontal distance it travels and the vertical distance it falls, I can find the total distance the aid will travel before it falls. I use pythagorean theorem to find this, and yield a value of approximately 296.5 m as the resultant distance. Did I do this correctly? This woul
a) If the supplies are released just as the plane flies directly overhead, will they touch down on land or in water? Justify your response with calculations that show exactly where the package will land.
b) When should the supplies be released so that they touch down very close to the centre of the dry patch of land? Answer in terms of distance from the target, not time.
Equations used:
Vertical distance = 1/2(acceleration)(delta t^2)
Horizontal distance = (horizontal velocity)(delta t)
c^2 = a^2 + b^2
a) I know that the vertical displacement is given, but the horizontal displacement is not. In order to find the horizontal component displacement, I would need to find the change in time time. I take into account that there is no initial velocity, so I can use the vertical distance = 1/2(acceleration)(delta t^2) rather than Vertical distance =(initial velocity)(delta t)+1/2(acceleration)(delta t^2) equation. Since I'm given vertical distance = 180m, I can isolate and solve for time which is approximately 6.1 seconds. Now I can find the horizontal distance that is traveled:
Horizontal distance = (horizontal velocity)(change in time) = (38.889 m/s after conversion)(6.1) = 235.6 m forward. Now that I have both the horizontal distance it travels and the vertical distance it falls, I can find the total distance the aid will travel before it falls. I use pythagorean theorem to find this, and yield a value of approximately 296.5 m as the resultant distance. Did I do this correctly? This woul