A plane is flying horizontally at a height of 500 m and a constant speed of 429 km/h when an object is projected downward at an initial speed of 35.0 m/s. if air resistance is neglected, the average vertical component of velocity between the object's release and its striking the ground is

1) The motion of an object under the action of gravity, when air resistance is neglected is called projectile motion. The path is a curve with the form of a parabola.

2) The equations that rule that motion are:

Horizontal speed:
Vx = Vox = constant.
In this case it is the same horizontal speed of the plane at the moment when the object was proyected. Vx = 429 km/h.

Vertical speed:

Vy = Voy + gt

d = yo - Voy×t - gt² / 2

Vy² = Voy² - 2gd

Since, you know Voy =0, g = 9.81 m/s², and d = 500m, you can use the last equation, leading to:

Vy² = 2(9.81m/s²)(500m) = 9,810 m²/s² ⇒ Vy = √(9,810 m²/s²) = 99 m/s

3) As per definition the average velocity is displacement / time, so you need to find the time to reach the ground.

You can use the formula Vy = Voy + gt
Clear t: t = (Vy - Voy) / g = (99 m/s - 0) / 9.81 m/s = 10.1s

Average vertical velocity = 500 m / 10.1s = 49.5 m/s.
You can convert to km/h: 49.5 m/s × 1km/1,000m × 3600s/h = 178 km/h

4) Conclusion: the average vertical component of velocity between the object's release and its striking the ground is 49.5 m/s or 178 km/h