The packet should be dropped 3636.55 horizontal meters before the target in order to hit the target
Trajectory is the path followed by a moving object when it is falling under gravitational force.
x = 90 t
y = -4.9 t²2 + 4500
t ≥ 0
y = 0
-4.9t² + 4500 = 0
4.9t²2 = 4500
[tex]\rm t = \sqrt {\dfrac{4500}{4.9}}[/tex]
t = 30.30 seconds
substituting the value to determine the horizontal distance
x = 120 * 30.30
x = 3636.55 meters
Therefore the packet should be dropped 3636.55 horizontal meters before the target in order to hit the target
The complete question is
a drone traveling horizontally at 120m s over a flat ground at an elevation of 4500 meters must drop an emergency package on a target on the ground the trajectory of the package is given by x 120t y 4.9t 2 4500 where the origin is the point on the ground directly beneath the drone at the moment of release. How many horizontal meters before the target should the package be released in order to hit the target? Round to the nearest meter.
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