Kinematics - projectile motion - time to maximum height?

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Kinematics -- projectile motion -- time to maximum height?

Homework Statement


The nozzle of a fire hose discharges water at a speed of 10 m/s. The nozzle is aimed straight up. How long does it take for a water drop to reach its maximum height?
Start with translating the question:
vi= 10 m/s
vf= 0 m/s
t=? s
a= -9.8 m/s^2
delta−x= ?

Homework Equations





The Attempt at a Solution



OK, so I'm assuming that solving this takes 2 steps. One entails finding the maximum height, then using a kinematic formula to find time. Is that logic sound?
 
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I tried this:
Vfy=Viy+Ay*t
0 m/s=10 m/s-9.8 m/s^2*t
-10 m/s=-9.8 m/s^2 *t
t=10 m/s / -9.8 m/s^2
t=1.02 s
 
1.02 s is not an option for this multiple choice problem, the closest is 1.0 s. Is that likely the answer then?

Edit: The question asks how long it takes to reach maximum height, so because it is a parabola, it stands to reason that 1.02s/2=0.51s is how long it takes to reach the maximum height. Right?
 
The data in the problem are given with two significant digits. Round of your result to t=1.0 s. .

What is parabola?

When does a body projected straight upward reach maximum height? What is its velocity at the apex?
What is the maximum height?

What would be the velocity at t=0.5 s? Would the water drop move upward or downward then? Does it reach the maximum height at t=0.5 s?

ehild
 
I don't know. I remember someone in class saying that projectile motion was a parabola...

t=1.0 s means what? The time it takes for the motion to finish? or the time at which the water reaches its maximum height?
 
The motion is not parabola. A graph of a function can be parabola.

You applied the equation Vfy=Viy+Ay*t. What do the letters mean? You substituted Vfy=0. If you throw up a pebble, will it reach the ground with zero velocity? Where is the velocity zero?
Just throw up something and see...

ehild