A hot air ballo0n is ascending straight up at a constant
speedof 7.0 m/s. When the balloon is 12.0 m above the ground, agun
fires a pellet straight up from ground level with an initialspped
of 30.0 m/s. Along the paths of the ballon and thepellet, there are
two places where each of them has the altitude atthe same time. How
far above ground level are theseplaces?

Lets call z to the vertical direction (z= is ground) . Then the positions of the balloon and the pellet, using the values of the velocities we are given, are:

[tex]z_b =\text{balloon position}\\z_p=\text{pellet position}\\z_b=(7\frac{m}{s})t\\z_p=30\frac{m}{s}(t-t_0)-\frac{g}{2}(t-t_0)^2\\g=9.8\frac{m}{s^2}[/tex]

How do we know the value of t₀? This is the time when the pellet is fired. At this time the pellet position is zero: its initial position. To calculate it we know that the pellet is fired when the ballon is in z = 12m. Then:

[tex]t_0=\frac{12}{7}s[/tex]

We need to know the when the z values of balloon and pellet is the same:

[tex]z_b=z_p\\(7\frac{m}{s})t =30\frac{m}{s}(t-\frac{12}{7}s)-\frac{g}{2}(t-\frac{12}{7}s)^2[/tex]

We need to find the roots of the quadratic equation. They are:

[tex]t_1=2.31s\\t_2=5.81[/tex]

To know the altitude where the to objects meet, we replace the time values:

[tex]z_1=16,17m\\z_2=40,67m[/tex]