A plane is flying horizontally at a height of 500 m and a constant speed of 429 km/h when an object is projected vertically 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

falls from 500 m with inital Vi = 35

v = Vi + a t so linear in time, average is half of (start + stop)
start v = 35

kinetic energy at ground = (1/2) m (35^2) + m g (500) = (1/2) m v^2

so
v^2 = 9.81 * 1000 + 1225
v = 105 m/s at ground
average v = (35 + 105)/2 = 70 m/s

To find the average vertical component of velocity between the object's release and striking the ground, we need to calculate the time it takes for the object to reach the ground and then use this time to find the average vertical component of velocity.

Let's start by calculating the time it takes for the object to reach the ground. We can use the kinematic equation:

h = ut + (1/2)gt²

where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is time.

Given:
Initial velocity (u) = 35.0 m/s (downward)
Height (h) = 500 m
Acceleration due to gravity (g) = 9.8 m/s² (downward)

Substituting the values into the equation:

500 = (35.0)t + (1/2)(9.8)t²

Simplifying the equation:

4.9t² + 35t - 500 = 0

To solve this quadratic equation for t, we can use the quadratic formula:

t = (-b ± √(b² - 4ac)) / 2a

where a = 4.9, b = 35, and c = -500.

Applying the values to the formula:

t = (-35 ± √(35² - 4(4.9)(-500))) / (2*4.9)

Calculating the values under the square root:

t = (-35 ± √(1225 + 9800)) / 9.8

t = (-35 ± √(11025)) / 9.8

t = (-35 ± 105) / 9.8

Solving for t:

t₁ = (-35 + 105) / 9.8 = 70 / 9.8 = 7.14 seconds (ignoring the negative result)

Now, we can calculate the average vertical component of velocity between the object's release and striking the ground.

Average velocity (V) = (Final velocity - Initial velocity) / 2

The final velocity is the vertical component of velocity when the object strikes the ground, which is the same as the plane's vertical speed.

Given:
Plane's vertical speed = 0 m/s (since it is flying horizontally)

Average velocity (V) = (0 - 35.0) / 2 = -17.5 m/s

Therefore, the average vertical component of velocity between the object's release and striking the ground is -17.5 m/s.

To find the average vertical component of velocity between the object's release and its striking the ground, we first need to determine the time it takes for the object to reach the ground.

We can use the equation for the vertical motion of the object:
h = ut + (1/2)gt^2

Here, h is the initial height (500 m), u is the initial vertical velocity (-35.0 m/s, since the object is projected downwards), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time.

Rearranging the equation, we get:
t = sqrt((2h)/g)

Plugging in the values, we have:
t = sqrt((2 * 500 m) / (-9.8 m/s^2))
t ≈ 10.1 s

Now that we know the time it takes for the object to reach the ground, we can find the average vertical component of velocity.

Average velocity is defined as the total displacement divided by the time taken. Since the object is projected downwards, its displacement is equal to the final velocity minus the initial velocity.

The final vertical velocity can be found using the equation:
v = u + gt

Here, v is the final vertical velocity, u is the initial vertical velocity, g is the acceleration due to gravity, and t is the time.

Plugging in the values, we have:
v = (-35.0 m/s) + (-9.8 m/s^2) * (10.1 s)
v ≈ -133.95 m/s

The negative sign indicates that the velocity is in the downward direction.

Finally, the average vertical component of velocity is given by:
average vertical velocity = total displacement / time taken

The total displacement is the difference between the initial and final velocity:
average vertical velocity = (-133.95 m/s - (-35.0 m/s)) / (10.1 s)

Simplifying this expression, we get:
average vertical velocity ≈ -9.46 m/s

Therefore, the average vertical component of velocity between the object's release and its striking the ground is approximately -9.46 m/s.