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
.
Answer:
graph A
Explanation:
the slope of the distance-time graph is speed, speed is a scalar (with magnitudes but no direction)
but the slope for the velocity time graph is acceleration, acceleration is vector quantity ( has magnitude and direction)
<span>1078 kgm / s would be the answer I hope this helps!!!</span>
Answer: R = 394.36ohm
Explanation: In a LR circuit, voltage for a resistor in function of time is given by:

ε is emf
L is indutance of inductor
R is resistance of resistor
After 4s, emf = 0.8*19, so:





R = 394.36
In this LR circuit, the resistance of the resistor is 394.36ohms.
DEFINITION:::::;;The type of reactions in which energy is releases to the environment are called Exothermic reactions.
EXAMPLE::: formation of carbon dioxide and urea formation are actually the examples of exothermic reaction..
Hope it helps