A plane traveling horizontally at 80m/s over at ground at an elevation of 3000m

Question

A plane traveling horizontally at 80m/s over

at ground at an elevation of 3000m releases

an emergency packet. The trajectory of the packet is given by

x = 80t; y = 4.9t^2 + 3000; for t greater than or equal to 0;

where the origin is the point on the ground directly beneath the plane at the moment of

release.

(1) Eliminate the parameter to nd a Cartesian equation for the trajectory of the packet.

(2) Graph the trajectory of the packet.

(3) How long does it take for the packet to land?

(4) Find the coordinates of the point where the packet lands.

(5) What was the length of the path taken by the packet?

Explanation / Answer

i think there is wrong with this question... as u can see.... y = 4.9t^2 + 3000 it as time goes...y keep increasing....packet never come down..... i think there should be -3000...

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