What is the Total Airtime of an Arrow Shot Straight Up at 20m/s?

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To determine the total airtime of an arrow shot straight up at 20 m/s, the relevant equations of motion must be applied. The initial velocity (u) is 20 m/s, and the acceleration due to gravity (a) is -9.81 m/s². The time to reach the maximum height can be calculated using the equation v = u + at, where the final velocity (v) at the peak is 0 m/s. Once the time to ascend is found, it should be doubled to account for the descent, yielding the total airtime. Understanding the correct formulas and the impact of gravity is crucial for solving this problem.
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Homework Statement


An arrow is straight up into the air at 20m/s what is the total airtime of the arrow


Homework Equations



i don't know what formula to use but i tried
F=MG

The Attempt at a Solution


F=Mg
 
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You need to look through the relevant equations for these kinds of problems and see what is applicable given the knowns in the question:

v = u + at

s = ut + 1/2at^2

s = vt - 1/2at^2

s =1/2(u + v)t

v^2 = u^2 + 2as

You are given the initial velocity (u) of the arrow, and you know (a) = 9.81 m/s^2

If something was thrown up in the air, what would happen to it eventually just before it starts to come down, what would its velocity (v) be then?

Also take into consideration the sign (+ or -) of gravity.
 
so do i get time first ? from v = u = at
 
v = u + at

What are you trying to solve for? Think of it conceptually, if an object is thrown up into the air, the total time up in the air is the time it spent going up and the time it spent coming down.
 
so i just find the time its going up and double it to make it going up and down

the problem is idk what formula 2 use
 
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