Step 1: Analyze the Given Conditions

The ball is projected at an initial velocity u = 60 m/s at an angle θ = 60° to the horizontal.

The angle of the velocity vector with the horizontal is initially 60°. Since we are looking for times t₁ and t₂ when the velocity vector again makes an angle of 60° with the horizontal, one of these times is clearly t₁ = 0 s (initial time).

Step 2: Velocity Components

The velocity components at any time t are given by:

v_x = u cos(θ)

v_y = u sin(θ) – g t

Where:

• u = 60 m/s
• θ = 60°
• g = 9.8 m/s² (acceleration due to gravity)

At time t, for the velocity vector to make an angle of 60° with the x-axis again:

tan(60°) = v_y / v_x

Step 3: Substitute Known Values

Since v_x = u cos(60°) = 60 × 0.5 = 30 m/s

And v_y = 60 × sin(60°) − 9.8t = 60 × 0.866 − 9.8t = 51.96 − 9.8t

Now, using the angle condition:

tan(60°) = (51.96 − 9.8t) / 30

⇒ √3 = (51.96 − 9.8t) / 30

⇒ 51.96 − 9.8t = 30√3

⇒ 9.8t = 51.96 − 30√3

Calculating:

30√3 ≈ 51.96 → So, t ≈ 0 (which we already knew as t₁)

For the second solution, use negative of the square root direction: −√3 = (51.96 − 9.8t) / 30

⇒ 51.96 − 9.8t = −30√3
⇒ 9.8t = 51.96 + 51.96
⇒ t = 103.92 / 9.8 ≈ 10.6 s