Finding time up and time down: Verticle Projectile

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The discussion revolves around calculating the time it takes for a vertical projectile to ascend and descend after being launched. The known variables include the height of the ball's flight and the mass of the ball, with initial calculations yielding a time down of 0.319 seconds, which was marked incorrect by the professor. Participants clarify that the time up and time down should be equal, leading to confusion about the correct application of physics formulas. There is a suggestion that the force of the spring may also influence the calculations. Ultimately, there is a consensus that the time to fall back to the top of the launcher is indeed 0.319 seconds, raising the possibility of a grading error by the professor.
Mirth
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[SOLVED] Finding time up and time down: Verticle 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...
 
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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...
 
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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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Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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