Help Physics (I think its free fall)

[coriimon]

A doctor, preparing to give a patient an injection, squirts a small amount of liquid straight upward from a syringe. If the liquid emerges with a speed of 1.5 m/s
how long does it take for it to return to the level of the syringe?
what is the maximum height of the liquid above the syringe?

The Attempt at a Solution

I got .30seconds to return to the level of the syringe
and the max height at around .11 meters


am I right?? just want to make sure

 

[Redbelly98]

If you show what equations you used and how you got those answers, it's easier to check your work.

 

[coriimon]

A doctor, preparing to give a patient an injection, squirts a small amount of liquid straight upward from a syringe. If the liquid emerges with a speed of 1.5 m/s
how long does it take for it to return to the level of the syringe?
what is the maximum height of the liquid above the syringe?

whoops sorry

I found the time it takes to return to the level of the syringe by using this equation

v=v(initial)=at
1.5m/s=0m/s+(9.81m/s^2)(t)
t=.15 seconds
to account for the time to return to the syringe I multiplied t by 2 so t=.30 seconds

to find the maximum height I used the equation
v^2=V(initial^2 +2a(x-x(initial))
2.25=19.62x
.11 m

 

[Redbelly98]

v=v(initial)=at
1.5m/s=0m/s+(9.81m/s^2)(t)
t=.15 seconds
to account for the time to return to the syringe I multiplied t by 2 so t=.30 seconds

I agree with both your final answers, but if I were grading this I might knock off a point here.

If you're finding the time it takes to reach the top of the trajectory, v is not 1.5 m/s there, as you've indicated. Also, the initial velocity isn't 0.

If you're finding the time it takes to drop from top of the trajectory, then the equation is okay ... provided you meant to use down as the + direction.

But it would only be a minor deduction. You pretty much got it on the nose.