1.2 A bullet is fired at a muzzle velocity of \( 125 \mathrm{~m} / \mathrm{s} \) at an angle of \( 25^{\circ} \) to the horizontal. Calculate the following: 1.2.1 The horizontal distance of the bullet after 8 seconds 1.2.2 The maximum height of the bullet

We are given:  • Muzzle velocity, u = 125 m/s  • Angle, θ = 25°  • Time for horizontal distance, t = 8 s  • Acceleration due to gravity, g = 9.8 m/s² Step 1. Find the initial velocity components:  • Horizontal component, uₓ = u cos θ = 125 cos 25°  • Vertical component, uᵧ = u sin θ = 125 sin 25° Using approximate values:  cos 25° ≈ 0.9063 and sin 25° ≈ 0.4226 Thus:  uₓ = 125 × 0.9063 ≈ 113.29 m/s,  uᵧ = 125 × 0.4226 ≈ 52.83 m/s. --------------------------------------------------------------------- 1.2.1 Horizontal Distance after 8 Seconds Since there is no horizontal acceleration (ignoring air resistance), the horizontal distance x is given by:  x = uₓ × t So:  x = 113.29 m/s × 8 s ≈ 906.32 m Thus, after 8 seconds the bullet is approximately 906 m away horizontally. --------------------------------------------------------------------- 1.2.2 Maximum Height of the Bullet The maximum height for a projectile (when starting from ground level) is given by:  H = (uᵧ)² / (2g) Substitute in:  H = (52.83 m/s)² / (2 × 9.8 m/s²) First, calculate (52.83)²:  (52.83)² ≈ 2791.4 m²/s² Then:  H ≈ 2791.4 / 19.6 ≈ 142.4 m Thus, the maximum height the bullet reaches is approximately 142 m. --------------------------------------------------------------------- Final Answers:  1.2.1 The horizontal distance after 8 seconds is approximately 906 m.  1.2.2 The maximum height reached by the bullet is approximately 142 m.