## Statement

difficulty

A rookie equilibrist is standing on a platform 12 meters above the ground. While practicing juggling with 2 balls, he stumbles and throws both balls at 9 m/s, however, he throws one up which we will call A and the other one down which we will call B. Considering that gravity is 10 m/s2, calculate

a) The time they are in the air.
b) Their velocity when they hit the ground.
c) The maximum height that ball A reaches.

## Solution

To solve this exercise, we will study each ball separately, since each one experiences a different motion:

Ball A. Upwards vertical launch.
Ball B. Downward vertical launch.

Question a)

Data

H = 12 m
v0 = 9 m/s
g = 10 m/s

Resolution

In both cases, to calculate the time that the balls stay in the air we must know at what time they touch the ground, that is, when their position y=0 m. Substituting in the equations of position for vertical motion:

Ball A

Ball B

Question b)

Data

H = 12 m
v0 = 9 m/s
g = 10 m/s
tA = 2.69 s
tB = 0.9 s

Resolution

Once we know the time that each ball takes to reach the ground, we can use that time to calculate their velocity at that moment by applying the formulas of vertical launch:

Ball A

Ball B

Question c)

Data

H = 12 m
v0 = 9 m/s
g = 10 m/s
tA = 2.69 s
tB = 0.9 s

Resolution

Ball A reaches its maximum height when its speed is 0 m/s. In the first place we will calculate the time it takes to reach this height:

Once we know the time, we can calculate the maximum height:

## Formulas worksheet

These are the main formulas that you must know to solve this exercise. If you are not clear about their meaning, we recommend you to check the theory in the corresponding sections. Furthermore, you will find in them, under the Formulas tab, the codes that will allow you to integrate these equations in external programs like Word or Mathematica.

Formulas
Related sections
$y=\mathrm{H}+{v}_{0}t-\frac{1}{2}g{t}^{2}$
$y=\mathrm{H}-{v}_{0}t-\frac{1}{2}g{t}^{2}$
$v={v}_{0}-g\cdot t$
$v=-{v}_{0}-g\cdot t$