## Related questions with answers

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

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

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

VerifiedAnswered 1 year ago

Answered 1 year ago

Step 1

1 of 6The graph of the trajectory of the packet

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

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

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