what is the magnitude of the total velocity of the package at the moment it is thrownas seen by an observer on


a plane is flying horizontally with speed 294 m/s at a height 2360 m above the ground, when a package is dropped from the plane. the acceleration of gravity is 9.8m/s.
horizontal distance from the relese point to the impact point is 6452.1534m
A second package is thrown downward from the plane with vertical speed V1=60 m/s.
--what is the magnitude of the total velocity of the package at the moment it is thrownas seen by an observer on the ground? answer in m/s
--what horizontal distance is traveled by this package? answer in m

Answer


1. magnitude of the total velocity
The package has an initial horizontal velocity and vertical velocity
Vih = 294 m/s
Viv = 60 m/s
So the magnitude of the total velocity is the hypothenuse:
Vi = sqrt(294^2 + 60^2) = 300 m/s
2. horizontal distance
1st checking with the first part of the text
d = 1/2 at^2
t = sqrt(2d/a) = sqrt(2x2360/9.8) = 21.946s
21.946s*294m/s = 6452. m
So there is no air resistance
2nd: because there is an initial vertical velocity of 60m/s the equation is
d = 1/2at^2 + vt (d,a and v are all positive since they are all oriented downward)
2360 = 9.8/2 t^2 + 60t
0 = 4.9t^2 + 60t - 2360
solve for t
t = 16.7 s
In that time the package will have traveled 4910 m (16.7s x 294m/s).


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