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 [itex] v_f^2-v_i^2=2a\Delta x [/itex] 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 [itex] v_i=0 [/itex] and [itex] \Delta x > 19.6 [/itex].

 

[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?

 

[HalfLostAndSearching]

B.) The time it takes for both balls to reach the ground can be found using the formula t=(Vf-Vi)/a. For the ball thrown downward, t=(0-14.7)/-9.8=1.5s. For the ball thrown upward, t=(0-14.7)/-9.8=1.5s. Therefore, the difference in time is 0s.

C.) The balls will have traveled the same distance horizontally since they are thrown with the same initial velocity and acceleration. Therefore, the distance between them will be 19.6m, the same as the initial height. However, since the balls are thrown at the same time, the difference in time will not affect their horizontal distance. Therefore, after 0.800s, the balls will still be 19.6m apart.

I would like to point out that there may be some inaccuracies in the given information. Firstly, the velocity of the balls should be measured in m/s, not just m. Also, the acceleration due to gravity should be -9.8 m/s^2, not just -9.8. These small details may not affect the overall conclusion, but it is important to be precise in scientific experiments and calculations.

Additionally, it is important to note that the equations used to solve for the velocities and time in the air assume that there is no air resistance. In reality, air resistance will affect the motion of the balls and may result in slightly different velocities and times. This is something that should be taken into consideration in further experiments.

Overall, the results of this experiment show that the initial velocity and acceleration do not affect the difference in time and horizontal distance between the two balls. This is because both balls are subject to the same acceleration due to gravity and are thrown at the same time, resulting in identical trajectories. This can be further tested by varying the initial velocities and observing the difference in time and distance between the balls.