Need help calculating maximum height

[farhan769]

Homework Statement

A doctor, preparing to give a patient an injection, squirts a small amount of liquid straight upward from a syringe.
a) If the liquid emerges with a speed of 1.3 , how long does it take for it to return to the level of the syringe?
b) What is the maximum height of the liquid above the syringe?

Homework Equations

y=(vi)(t)-.5(g)(t^2)
d=vt
i'm really not sure what equation to use

The Attempt at a Solution

for part a i use y=(vi)(t)-.5(g)(t^2)
0=1.3t-.5(-9.8)(t^2)
t=.27s

I'm really not sure about part b, could someone help me with it
thanks

 

[LowlyPion]

OK. We'll go with 1.3 m/s (you must include units ... it matters).

The easy way is to observe that

v = a*t

so the time to max height will be given by

1.3 = 9.8*t or t = 1.3/9.8.

Since total time is time up + time down ... double that giving 1.3*2/9.8

Since y = 1/2*g*t² then just use the time found to max height from before of 1.3/9.8

 

[nlightner]

I would suggest using the equation for maximum height, which is h = (vi)^2/2g. In this case, vi=1.3 and g=9.8. Plugging these values into the equation, we get h = (1.3)^2/2(9.8) = 0.0877 meters. This is the maximum height that the liquid will reach above the syringe.