Finding time up and time down: Verticle Projectile

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Homework Help Overview

The discussion revolves around a vertical projectile problem involving a launcher that propels a ball upward and measures the time taken for the ball to return to the launcher. The known parameters include the height of the ball's flight and the mass of the ball, with participants exploring the calculations for time taken during ascent and descent.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss various methods to calculate the time of ascent and descent, questioning the validity of the original poster's calculations. There is a focus on whether the time down is equal to the time up and considerations regarding the influence of the spring force.

Discussion Status

Some participants are clarifying the specific time being calculated (to the top of the launcher versus the floor) and expressing uncertainty about the professor's feedback on the calculations. There is a suggestion that the professor may have made an error in marking the original poster's work.

Contextual Notes

The original poster mentions confusion regarding the impact of the spring force on the calculations, indicating potential gaps in the problem setup or assumptions being made.

Mirth
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[SOLVED] Finding time up and time down: vertical Projectile

Hey guys, so I have a take home lab that is just murdering me...

Imagine a vertical projectile launcher that sits on the ground, that shoots a ball up in the air and lands back in the projectile.

The knowns are:
distance from the top of the projectile launcher to the top of the ball's flight = 0.5m
distance of the compressed spring = .033m
and mass of ball = .00965kg

Now, I thought time down could be gotten using Sqrt(2 * d/ g), which would be Sqrt(2 * .5m/9.8) = .319 seconds, but the professor marked that wrong.

Other than that, I know that V_i = 0 when the ball is at the top of it's flight. So I decided to get V_f by using V_f = Sqrt(2 * g * d), which is Sqrt(2 * 9.8 * .5m) = 3.13 seconds. But I don't know how I can get time down using those two.

Also, I "figure" time down would be the same as time up. With that said, if the initial velocity is 3.13m/s and the final velocity is 0m/s for the time up, then V_f - V_i / g
= 0 - 3.13 / 9.8 = 3.19s . That's what I got for my time down on my paper, so if time up is = to time down, and I got my time down marked wrong when it was .319, then both aren't .319 :(

Edit: Err, but then wouldn't the force of the spring be a factor? I'm so confused...
 
Last edited:
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Are you trying to find the time it takes the projectile to fall back down to the top of the launcher or are you trying to find the time it takes to fall back to the foor?

How high is this launcher? If it is significantly high, the time it time it takes to hit the floor may be significantly longer.
 
G01 said:
Are you trying to find the time it takes the projectile to fall back down to the top of the launcher or are you trying to find the time it takes to fall back to the foor?

How high is this launcher? If it is significantly high, the time it time it takes to hit the floor may be significantly longer.

Fall back to the top of the launcher, not the floor, yup.

Someone sort of helped me for the time down, and he also got .319s, so I think my professor may have marked it wrong by accident...
 
Last edited:
Mirth said:
Fall back to the top of the launcher, not the floor, yup.

Someone sort of helped me for the time down, and he also got .319s, so I think my professor may have marked it wrong by accident...

Quite possible. Professors are human too, despite evidence suggesting otherwise:smile:
 

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