To find the time of flight, maximum height, and horizontal range of the object, we can use the kinematic equations of motion.

1. Time of flight (T):
We know that the vertical component of the initial velocity is given by v₀y = v₀sin(θ), where v₀ is the initial velocity and θ is the launch angle. In this case, v₀ = 30 m/s and θ = 60 degrees.

Using the equation of motion, Δy = v₀yt + 0.5at², where Δy is the vertical displacement, a is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time of flight, we can solve for t.

At the maximum height, the vertical displacement is zero since the object reaches its highest point. Therefore, Δy = 0. Substituting the values into the equation, we get:
0 = v₀yt - 0.5gt²

Simplifying, we have:
0.5gt² = v₀yt

Dividing both sides by t and rearranging the equation, we get:
t = (2v₀y) / g

Substituting the values as mentioned above, we get:
t = (2 * 30 * sin(60)) / 9.8 ≈ 3.07 seconds

Therefore, the time of flight is approximately 3.07 seconds.

2. Maximum Height (H):
The maximum height can be found using the equation, Δy = v₀yt - 0.5gt².

Substituting the known values, we have:
H = (30 * sin(60) * 3.07) - 0.5 * 9.8 * (3.07)²

Simplifying, we get:
H ≈ 44.1 meters

Therefore, the maximum height is approximately 44.1 meters.

3. Horizontal Range (R):
The horizontal range can be found using the equation R = v₀xt.

The horizontal component of the initial velocity is given by v₀x = v₀cos(θ), where v₀ = 30 m/s and θ = 60 degrees.

Using the equation, R = v₀xt, and substituting the known values, we have:
R = (30 * cos(60)) * (3.07)

Simplifying, we get:
R ≈ 88.2 meters

Therefore, the horizontal range is approximately 88.2 meters.

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