Correct Answer: Option A

Explanation

The motion of the particle can be described using the equation of displacement under constant acceleration:  
\(s = ut + \frac{1}{2}at^2\)
where:  
\( s \) is the displacement (here, \( s = -45 \) m, taking upward as positive and the ground as the final position),  
\( u = 40 \) m/s (initial velocity upward), \( a = -g = -10 \) m/s² (acceleration due to gravity, approximating \( g = 10 \) m/s²),  
- \( t \) is the time.  
\(-45 = 40t + \frac{1}{2}(-10)t^2\)
\(-45 = 40t - 5t^2\)
Rearrange into standard quadratic form:  
\(5t^2 - 40t - 45 = 0\)
Divide through by 5:  
\(t^2 - 8t - 9 = 0\)

Solve using the quadratic formula:  
\(t = \frac{8 \pm \sqrt{(-8)^2 - 4(1)(-9)}}{2} = \frac{8 \pm \sqrt{64 + 36}}{2} = \frac{8 \pm \sqrt{100}}{2} = \frac{8 \pm 10}{2}\) 
The solutions are:  
\(t = \frac{8 + 10}{2} = 9 \text{ s}\)
\(t = \frac{8 - 10}{2} = -1 \text{ s (discard, as time cannot be negative).}\)

Thus, the time taken to hit the ground is 9 seconds.

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