Initially (at time t = 0) a particle is moving vertically at 5.8 m/s and horizontally at 0 m/s. Its horizontal acceleration is 2.1 m/s2.

At what time will the particle be traveling at 54� with respect to the horizontal? The acceleration due to gravity is 9.8 m/s2.

The question here is what force other than grvity is the particle being subjected to? You stated a horizontal acceleration, but what about vertical? is the vertical acceleration zero?

The vertical acceleration is 9.8 m/s^2 because of gravity. So in the negative y direction.

I think it's saying only gravity is affecting the particle on the y-axis.

To find the time at which the particle will be traveling at an angle of 54 degrees with respect to the horizontal, we can break down the motion of the particle into its horizontal and vertical components.

Given:
Initial vertical velocity (Vy0) = 5.8 m/s
Initial horizontal velocity (Vx0) = 0 m/s
Horizontal acceleration (ax) = 2.1 m/s²
Angle with respect to the horizontal (θ) = 54 degrees
Acceleration due to gravity (g) = 9.8 m/s²

First, we need to find the vertical component of the velocity (Vy) when the particle reaches an angle of 54 degrees.

Using the basic trigonometric relationship, we can find Vy:
Vy = Vy0 + gt

Since the particle is moving vertically, the horizontal velocity (Vx) remains constant at 0 m/s.
Hence, the angle between the resultant velocity vector and the horizontal axis is the same as the angle between the vertical velocity vector and the resultant velocity vector.

We can find the resultant velocity (V) using the pythagorean theorem:
V = sqrt(Vx^2 + Vy^2)

To create a triangle to simplify the calculation, we can find the horizontal component of the resultant velocity (Vx) using trigonometry:
Vx = cos(θ) * V

Now we have the horizontal component of the velocity (Vx) and the horizontal acceleration (ax), so we can determine the time it takes for the particle to reach an angle of 54 degrees with respect to the horizontal using the equation:
Vx = ax * t

Solving for t gives us:
t = Vx / ax

Let's substitute the known values into the equations to find the time (t):

1. Calculate the vertical component of the velocity (Vy):
Vy = Vy0 + gt
Vy = 5.8 m/s + (9.8 m/s²) * t

2. Calculate the horizontal component of the resultant velocity (Vx):
V = sqrt(Vx^2 + Vy^2)
Vx = cos(θ) * V
Vx = cos(54 degrees) * V

3. Calculate the resultant velocity (V):
V = sqrt(Vx^2 + Vy^2)
V = sqrt((cos(54 degrees) * V)^2 + (5.8 m/s + (9.8 m/s²) * t)^2)

4. Calculate the time at which the particle will be traveling at an angle of 54 degrees with respect to the horizontal:
t = Vx / ax
t = (cos(54 degrees) * V) / 2.1 m/s²

By substituting the known values into the equations and solving for t, we can determine the time at which the particle will be traveling at an angle of 54 degrees with respect to the horizontal.