How many seconds before he is directly above the platform must he release the package and what must be the distance measured in the horizontal direction separating the plane and the platform?
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The pilot is flying at a velocity 300 km/h horizontally at a height 300 m above the ground. The velocity of the platform is 30 km/h horizontally. Relative to the platform the pilot is flying at 270 km/h.
When the pilot drops the packet, it has a horizontal velocity of 270 km/h relative to the platform. The packet does not have any vertical velocity. The time taken by the packet to fall 300 m is t that follows the relation .300 = (1/2)*9.8*t^2
=> .3*2/9.8 = t^2
=> t = .2474 s
In .2474 s the horizontal distance moved by the packet relative to the platform is 18.55 m
The pilot has to drop the packet 0.2474 s before he is directly above the platform. The horizontal distance separating the two is 18.55 m.
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