Mathphys
second part says

what is the horizontal component of the velocity just before it hits ?

third part is

what is the vertical component of the velocity just before it hits?

everything is answered in m/s (but im pretty sure thats for like literally everything i don't know)

270 m

45 m/s

59 m/s

Explanation:

Part 1.

First, to find where the package lands, we must find how long it takes to fall.

Given (in the y direction, taking down to be positive):

Δy = 175 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(175 m) = (0 m/s) t + ½ (9.8 m/s²) t²

t = 5.98 s

Next, find the distance traveled in that time.

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (45 m/s) (5.98 s) + ½ (0 m/s²) (5.98 s)²

Δx = 270 m

Part 2.

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

Find: v

v² = v₀² + 2aΔx

v = (45 m/s)² + 2 (0 m/s²) Δx

v = 45 m/s

Part 3.

Given (in the y direction, taking down to be positive):

Δy = 175 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (9.8 m/s²) (175 m)

v = 59 m/s

nah.   i'm just sleepy.

the bal will not change its velocity.