Question

A plane traveling horizontally at 98 ​m/s over a flat ground at an elevation of m must drop an emergency packet on a target on the ground. The trajectory of the packet is given by xequals98​t, yequalsminus4.9tsquaredplus3500​, for tgreater than or equals​0, where the origin is the point on the ground directly beneath the plane at the moment of the release. How many horizontal meters before the target should the packet be released in order to hit the​ target?

Answer 1

The packet should be released approximately 2169 meters before the target, to hit it

Explanation:

Trajectory of the packet:

x = 98*t

y = -4.9*
`t^(2)` + 3500

t ≥ 0

If we do y = 0 we can determine t for when the packet hit ground (the time between drop and hit)

y = -4.9*
`t^(2)` + 3500

0 = -4.9*
`t^(2)` + 3500

t =
`\sqrt{(3500)/(4.9) }`

Then using the equation of x, we can calculate how far in horizontal direction the packet must be drop to hit the target in the calculated t .

x = 98*t

x= 98 *
`\sqrt{(3500)/(4.9) }`

x= 2169

The packet should be released approximately 2169 meters before the target, to hit it