We<span> use </span>inequalities<span> when there is a range of possible answers for a situation. ... </span>inequalities—inequalities<span> that can be written in the form of a linear </span>equation. ... the bounded region, and anypoint<span> within this region </span>satisfies<span> the </span>inequality<span> x ≥ -</span><span>2. ... </span>All<span> of the </span>points<span> under the line are shaded; this is the range of </span>points<span> where the ...</span>
gas molecules having at least one oxygen atom
The person's horizontal position is given by

and the time it takes for him to travel 56.6 m is

so your first computed time is the correct one.
The question requires a bit of careful reading, and I think there may be a mistake in the problem. The person's vertical velocity
at time
is

which tells us that he would reach the ground at about
. In this time, he would have traveled

But we're told that he is caught by a net at 56.6 m, which would mean that the net cannot have been placed at the same height from which he was launched. However, it's possible that the moment at which he was launched doesn't refer to the moment the cannon went off, but rather the moment at which the person left the muzzle of the cannon a fraction of a second after the cannon was set off. After this time, the person's initial vertical velocity
would have been a bit smaller than
.
Complete Question
Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is the nail compressed if it is 2.50 mm in diameter and 6.00-cm long.What Average force is excreted on the Nail
Answer:

Explanation:
From the question we are told that:
Mass 
Initial Velocity 
Distance 
Diameter 
Length 
Generally the equation for Force is mathematically given by



Answer:
The work done by the applied force is 259.22 J.
Explanation:
The work done by the applied force is given by:

Where:
F: is the applied horizontal force = 108.915 N
d: is the distance = 2.38 m
Hence, the work is:

Therefore, the work done by the applied force is 259.22 J.
I hope it helps you!