Comparing Balls' Velocity & Time in the Air

[Chasezap]

Homework Statement

Two students conduct physics experiments on a balcony 19.6 m above the street. One student throws a ball vertically downward at 14.7 m/s. At the same instant the second student throws a ball vertically upward at the sam speed. The second ball follows the same path as the first ball on its way down.
ΔX=19.6
Vi=14.7
a=-9.8

Homework Equations

A.) What are the velocities of each ball at the instant before they strike the ground? B.) What is the difference in the time the balls spend in the air? C.) How far apart are the balls 0.800s after they are thrown?

The Attempt at a Solution

A.) Vf^2=14.7^2+2(-9.8)(19.6)
Vf^2=-168.07
The square root of a negative number is not a real number obviously so not sure what to do here.

 

[ShayanJ]

For the ball thrown downward, you can use v_f^2-v_i^2=2a\Delta x directly, but not for the ball thrown upward. You should at first find out how much the ball goes up to stop(find its peak), and then use the peak as the first point which of course means v_i=0 and \Delta x > 19.6.

 

[CWatters]

I would have a think about the problem a bit more before tying to solve part a). It might be easier than you think. What do you know about objects thrown vertically upwards?

 

[nasu]

Homework Statement

3. The Attempt at a Solution
A.) Vf^2=14.7^2+2(-9.8)(19.6)
Vf^2=-168.07
The square root of a negative number is not a real number obviously so not sure what to do here.

Why did you put a negative sign in front of the value for acceleration?
If you consider acceleration negative this means you consider "up" as positive. What will be the sign of the displacement (the 19.6) in this case?