A projectile is launched from the edge of a cliff with a velocity of 15.0 m/s at an angle of 30.0 degrees above the horizontal. Ignoring air resistance, what is the velocity of the projectile 2.00 seconds later?

To solve this problem, we need to break down the initial velocity into its horizontal and vertical components, and then use the equations of motion to find the velocity at the given time.

Step 1: Find the horizontal and vertical components of the initial velocity. The horizontal component (Vx) is given by Vcos(θ) and the vertical component (Vy) is given by Vsin(θ).

Given that V = 15.0 m/s and θ = 30.0 degrees, we find:

Vx = 15.0 m/s * cos(30.0) = 12.99 m/s Vy = 15.0 m/s * sin(30.0) = 7.5 m/s

Step 2: Find the vertical velocity at the given time. The vertical velocity at a given time t is given by Vy - g*t, where g is the acceleration due to gravity (9.8 m/s²).

Given that t = 2.00 s, we find:

Vy(t) = 7.5 m/s - 9.8 m/s² * 2.00 s = -12.6 m/s

Step 3: Find the resultant velocity at the given time. The resultant velocity is given by the vector sum of the horizontal and vertical velocities. Its magnitude is given by sqrt(Vx² + Vy(t)²) and its direction is given by atan(Vy(t)/Vx).

We find:

V(t) = sqrt((12.99 m/s)² + (-12.6 m/s)²) = 18.0 m/s θ(t) = atan(-12.6 m/s / 12.99 m/s) = -44.5 degrees

So, the velocity of the projectile 2.00 seconds later is 18.0 m/s at an angle of -44.5 degrees below the horizontal.