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Answer:
The horizontal distance traveled by the projectile is 373.36 m.
Explanation:
Given;
angle of projection; θ = 30⁰
initial velocity of the projectile, u = 65 m/s
The horizontal distance traveled by the projectile or range of the projectile is calculated as;
[tex]R = \frac{u^2 \times sin(2\theta)}{g} \\\\R = \frac{65^2 \times sin(2\times 30^0)}{9.8}\\\\R = \frac{4225\times sin(60)}{9.8} \\\\R = 373.36 \ m[/tex]
Therefore, the horizontal distance traveled by the projectile is 373.36 m.