An alaskan rescue plane traveling 48 m/s drops a package of emergency rations from a height of 165 m to a stranded party of explorers. the acceleration of gravity is 9.8 m/s 2 . where does the package strike the ground relative to the point directly below where it was released? answer in units of m.

what is the horizontal component of the velocity just before it hits?
answer in units of m/s.

Package strikes ground 278m from where it was released.

Horizontal component of velocity is still 48m/s.

Explanation:

To figure out the horizontal distance travelled (the range), we need to know how long the package was traveling through the air. To figure that out, we use the vertical information:

Given the following vertical information:

d = 165

a = 9.8

vi = 0

t = ?

We know the package was airborne for 5.8 seconds. Now we use that when considering all the horizontal stuff. Given:

t = 5.8

v = 48

d = ?

The horizontal component of the velocity is always the same because there are no horizontal forces acting on the package (assuming we're ignoring drag here).

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.