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Question

A plane traveling horizontally at 80 m/s} over flat ground at an elevation of 3000 m releases an emergency packet. The trajectory of the packet is given by

x=80t,y=4.9t2+3000, for t0,x=80 t, \quad y=-4.9 t^2+3000, \quad \text { for } t \geq 0,

where the origin is the point on the ground directly beneath the plane at the moment of the release. Graph the trajectory of the packet and find the coordinates of the point where the packet lands.

Solution

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Answered 1 year ago
Answered 1 year ago
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The graph of the trajectory of the packet

x(t)=80t,y(t)=4.9t2+3000,t0x(t)=80t\,\, , \,\, y(t)=-4.9t^2+3000\,\, , \,\, t\geq 0

is given bellow.,The Wolfram Mathematica code we used to generate the graph is also given.

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