(a) The time taken for the projectile to reach the maximum height is 32.65 s.
(b) The horizontal range of the projectile is 9,049.1 m.
The given parameters:
The time taken for the projectile to reach the maximum height is calculated as follows;
[tex]v_f = u- gt\\\\0 = 320 - 9.8t\\\\9.8t = 320\\\\t = \frac{320}{9.8} \\\\t = 32.65 \ s[/tex]
The horizontal range of the projectile is calculated as follows;
[tex]R = \frac{u^2 sin(2\theta)}{g} \\\\R = \frac{320^2 \times sin(2\times 30)}{9.8} \\\\R = 9,049.1 \ m[/tex]
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