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

Answer: 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

Explanation:

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