Aplane is flying horizontally with speed 107 m/s at a height 6680 m above the ground, when a package is dropped from the plane. the acceleration of gravity is 9.8 m/s 2 . neglecting air resistance, when the package hits the ground, the plane will be 1. behind the package. 2. directly above the package. 3. ahead of the package. 014 (part 2 of 4) 10.0 points what is the horizontal distance from the release point to the impact point? answer in units of m. 015 (part 3 of 4) 10.0 points a second package is thrown downward from the plane with a vertical speed v1 = 77 m/s. what is the magnitude of the total velocity of the package at the moment it is thrown as seen by an observer on the ground? answer in units of m/s. 016 (part 4 of 4) 10.0 points what horizontal distance is traveled by this package? answer in units of m

what we know is the horizontal velocity of 107m/s, the vertical distance of 6680m, and both vertical and horizontal accelerations of 0(horizontal) and -9.8m/s2(vertical). we are in this case trying to find the horizontal distance so we can only use 107 for velocity and 0 for the acceleration.

first we will find time using t= square root of (2 x vertical distance) divided by 9.8 which should give you 36.922 sec for both the horizontal and vertical times.

now we can solve for the horizontal distance (dx) using x=Vi(t)+.5(a)(t^2)   so this now looks like x=107 x 36.922 + .5 x 0 x 36.922 which can be reduced to               x=107x36.922 which equals 3950.654m

so the correct choice should be 1

hope this helps if it isnt too late :)

a charge 3q is at the origin, and a charge −2q is on the positive x axis at x=a. part a where on the x-axis would you place a third charge so it would experience no net electric force? enter an expression for the exact answer, which will involve a square-root sign. (do not enter an approximate decimal answer.)