A projectile is launched with an initial speed of 30 m/s atan angle of 60° above the horizontal. What are the (a) magni-tude and (b) angle of its velocity 2.0 s after launch, and (c) is theangle above or below the horizontal? What are the (d) magni-tude and (e) angle of its velocity 5.0 s after launch, and (f) is theangle above or below the horizontal?

Sure, let's solve this step by step.

First, we need to find the initial horizontal (Vox) and vertical (Voy) components of the velocity. We can use the following equations:

Vox = Vo * cos(θ) Voy = Vo * sin(θ)

where Vo is the initial speed (30 m/s) and θ is the angle (60°).

(a) and (b) 2.0 s after launch:

1. Calculate Vox and Voy: Vox = 30 m/s * cos(60°) = 15 m/s Voy = 30 m/s * sin(60°) = 25.98 m/s

2. The vertical velocity 2.0 s after launch (Vy) can be calculated using the equation: Vy = Voy - g*t where g is the acceleration due to gravity (9.8 m/s²) and t is the time (2.0 s). Vy = 25.98 m/s - 9.8 m/s² * 2.0 s = 6.38 m/s

3. The magnitude of the velocity (V) can be calculated using the Pythagorean theorem: V = sqrt(Vox² + Vy²) V = sqrt((15 m/s)² + (6.38 m/s)²) = 16.28 m/s

4. The angle (α) can be calculated using the arctan function: α = arctan(Vy/Vox) α = arctan(6.38 m/s / 15 m/s) = 23.19°

(c) The angle is above the horizontal because the projectile is still going upwards.

(d) and (e) 5.0 s after launch:

1. The vertical velocity 5.0 s after launch (Vy) can be calculated using the equation: Vy = Voy - g*t Vy = 25.98 m/s - 9.8 m/s² * 5.0 s = -23.02 m/s

2. The magnitude of the velocity (V) can be calculated using the Pythagorean theorem: V = sqrt(Vox² + Vy²) V = sqrt((15 m/s)² + (-23.02 m/s)²) = 27.32 m/s

3. The angle (α) can be calculated using the arctan function: α = arctan(Vy/Vox) α = arctan(-23.02 m/s / 15 m/s) = -56.31°

(f) The angle is below the horizontal because the projectile is now going downwards.