Answer:

32.74 m/s upwards.

Explanation:

To find the vertical position and velocity of a volleyball after 3.0 seconds, we can use the equations of motion for projectile motion.

Given: Initial velocity (v0) = 5.0 m/s Angle of projection (θ) = 45° Time (t) = 3.0 sec

First, we can break down the initial velocity into its horizontal and vertical components. The horizontal velocity (v0x) is given by: v0x = v0 * cos(θ)

The vertical velocity (v0y) is given by: v0y = v0 * sin(θ)

Substituting the given values: v0x = 5.0 * cos(45°) ≈ 3.54 m/s v0y = 5.0 * sin(45°) ≈ 3.54 m/s

Next, we can calculate the vertical position (y) after 3.0 seconds using the following equation: y = v0y * t + 0.5 * g * t^2

Where g is the acceleration due to gravity, which is approximately 9.8 m/s^2 on Earth.

Substituting the given values: y = 3.54 * 3 + 0.5 * 9.8 * 3^2 y ≈ 16.61 m

So, the vertical position of the volleyball after 3.0 seconds is approximately 16.61 meters above the initial point.

Finally, we can calculate the vertical velocity (vy) after 3.0 seconds using the following equation: vy = v0y + g * t

Substituting the given values: vy = 3.54 + 9.8 * 3 vy ≈ 32.74 m/s

So, the vertical velocity of the volleyball after 3.0 seconds is approximately 32.74 m/s upwards.

*KirkTheGreat

*AskAndYou'llRecieve

*OneFollowIsEnoughToSustainMyNeeds