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

  • Thread starter Thread starter antphany
  • Start date Start date
  • Tags Tags
    Air Time
AI Thread Summary
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.
antphany
Messages
11
Reaction score
0

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
 
Physics news on Phys.org
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
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top