To find out when the object hits the ground, we set the height function \( h(x) \) equal to zero. The function is given by:
\[ h(x) = -5(x+1)(x-9) \]
Setting \( h(x) \) to zero yields:
\[ -5(x+1)(x-9) = 0 \]
Since \(-5\) does not affect the roots, we can simplify the equation by dividing both sides by \(-5\):
\[ (x+1)(x-9) = 0 \]
Now, we can solve for \(x\) by setting each factor equal to zero:
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\( x + 1 = 0 \) \[ x = -1 \]
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\( x - 9 = 0 \) \[ x = 9 \]
The two solutions are \( x = -1 \) seconds and \( x = 9 \) seconds. Since negative time doesn't make sense in this context, we discard \( x = -1 \).
Therefore, the object will hit the ground \( 9 \) seconds after launch.
Thus, the answer is:
\[ \boxed{9} \text{ seconds} \]