How Long Does a Projectile Stay in Air When Launched Vertically at 10 m/s?

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Homework Help Overview

The discussion revolves around a projectile launched vertically at an initial velocity of 10 m/s. Participants are exploring the time the projectile remains in the air, considering the effects of gravity and the symmetry of the projectile's motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using kinematic equations to determine the time of flight, questioning how to account for the ascent and descent phases of the projectile's motion. There is an exploration of the relationship between the time to reach maximum height and the total time in the air.

Discussion Status

The discussion is active with various approaches being considered. Some participants suggest using specific equations to find the time to maximum height and then doubling that time for the total flight duration. Others are clarifying the implications of the projectile returning to its starting position and how that relates to displacement and time.

Contextual Notes

Participants note the assumption of constant acceleration due to gravity and the lack of specific distance information in the problem statement. There is also mention of the need to clarify the relationship between ascent and descent times.

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Homework Statement


A projectile launched straight up at 10 m/s will be in the air for:

A) 1 s

B) 1.4 s

C) 2 s

D) 5 s


Homework Equations


Not sure.


The Attempt at a Solution



I'm honestly not sure what to make of this problem. Since it's freefall, it sounds like we could assume the acceleration to be -10 m/s2. V0 would be equal to 10 m/s and V would equal 0 if it reaches a maximum height. If we need to find t and since we aren't given d, can we use the equation:

v = v0+at?
 
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There are a couple ways for this problem. That equation will work to find the time it takes to reach its maximum height, but keep in mind that the time in the air will be the time it takes to reach that height and then return to the starting height. Since the path is symmetric (i.e. same motion up and down), how does that time compare to the total time?
 
If the equation in my first post will help to find the total amount of time that it takes for the ball to complete the height, then would we need to find half the amount of time it would take for the ball to reach maximum height and then double that time to find the total time to reach the point (both in reaching max height and falling back down). Is that right?
 
I don't quite understand what you said.

The equation finds the time taken for the projectile to come to rest.As we know,the projectile will come to rest at maximum height and again start fall down.
So now we know the time taken to reach maximum height.Can you find the time taken to fall down?
There are two ways.
 
Time of Flight = 2u(sinΘ)/g

Also what is implied when an object is projected and returns to it's starting position wrt to displacement...you can also form an equation with this involving time
 
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