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).

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y