An Alaskan rescue plane traveling 39 m/s drops a package of emergency rations from a height of 198 m to a stranded party of 3 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?

What is the vertical component of the velocity just before it hits? (Choose upward as the
positive vertical direction)

y = 0m

y0 = 166m

v0y = 0 m/s

g = 9.8 m/s^2

t = ?

Solve for t:

y = y0 + v0y*t - (0.5)gt^2

0 = 166 - (0.5)(9.8)t^2

t = 5.82 s

Now, using time, we can solve for the range using the equation:

x = vx(t)

x = (40)(5.82)

x = 232.8 m

The impact horizontal component of velocity will be 40 m/s as velocity in terms of x is always constant. To find the impact vertical component of velocity, we use the equation:

v = v0y - gt

v = 0 - (9.8)(5.82)

v = -57.04 m/s

i still. don't get the answer is help me pls

b.

explanation:

10.0

c. magnitude and direction