< b="" />Overview: < />
In this problem, a particle is projected with a velocity of 80 feet/second and after 5 seconds, it moves at right angles to the direction of projection. We need to determine the angle of projection to the horizontal.

< b="" />Given: < />
Initial velocity (u) = 80 feet/second
Time (t) = 5 seconds

< b="" />Approach: < />
To find the angle of projection, we can use the concept of vectors and trigonometry. We'll break down the initial velocity into its horizontal and vertical components and then analyze the particle's motion at the given time.

< b="" />Step 1: Finding the Horizontal and Vertical Components of Velocity: < />
The horizontal component of velocity (u_x) remains constant throughout the motion, while the vertical component of velocity (u_y) changes due to the gravitational acceleration.

Horizontal component (u_x) = u * cos(θ)
Vertical component (u_y) = u * sin(θ)

Here, θ represents the angle of projection to the horizontal.

< b="" />Step 2: Analyzing the Particle's Motion: < />
After 5 seconds, the particle moves at right angles to the direction of projection. This means that its vertical component of velocity becomes zero, and only the horizontal component of velocity remains.

Using the equation of motion for vertical displacement (s_y), we can find the time at which the particle reaches the highest point of its trajectory.

s_y = u_y * t - (1/2) * g * t^2
0 = u * sin(θ) * t - (1/2) * g * t^2

Simplifying the equation, we get:
(1/2) * g * t^2 = u * sin(θ) * t
2g = u * sin(θ)
sin(θ) = 2g / u

Now, we can find the angle of projection (θ) using the inverse sine function (sin^-1):
θ = sin^-1(2g / u)

< b="" />Step 3: Calculating the Angle of Projection: < />
Substituting the given values, we have:
θ = sin^-1(2 * 32.2 / 80)
θ = sin^-1(0.806)

Using a scientific calculator, we find:
θ ≈ 53.13 degrees

< b="" />Answer: < />
Therefore, the angle of projection to the horizontal is approximately 53.13 degrees.

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