A ball is dropped from a balloon going up at a speed of \[7m/s\]. If the balloon was at height $60m$ at the time of dropping the ball, how long would it take to reach the ground?

Hint We have already been given the height and the initial velocity of the balloon, so in Newton's third equation of motion substitute these values and solve to obtain the value of time.
Formula used:
$S = ut + \dfrac{1}{2}a{t^2}$ where $S$ is the distance travelled, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time.

Complete step by step answer
This problem can be solved by implementing the equations of motions.
These equations of motions describe the behavior of a physical system based on their initial velocity, final velocity, acceleration and time.
Motion can be classified into two basic types- dynamics and kinematics.
In dynamics the forces and energies of the particles are taken into account. Whereas in kinematics only the position and time of the particle are taken into consideration.
The three main equations of motions are
$v = u + at$
${v^2} = {u^2} + 2aS$
$S = ut + \dfrac{1}{2}a{t^2}$
Where $v$ is the final velocity, $u$ is the initial velocity, $t$ is the time taken, $a$ is the acceleration and $S$ is the displacement of the body.
Now as the balloon is going up, so the initial velocity of the balloon is given as
$u = - 7m/s$
At height $60m$the ball is dropped
So $S = 60m$
Using the third equation of motion we get,
$
  S = ut + \dfrac{1}{2}a{t^2} \\
   \Rightarrow 60 = - 7t + \dfrac{1}{2} \times 9.8 \times {t^2} \\
$
Where $a = g = 9.8m/{s^2}$ which is the acceleration due to gravity and $t$ is the time taken by the ball to reach the ground.
So, $4.9{t^2} - 7t - 60 = 0$
Solving this quadratic equation using Shreedhara Acharya’s formula, we get
$t = 4.28s$

Thus, the ball takes $t = 4.28s$ to reach the ground.

Note The equations of motion give us a comprehensive idea about the behavior of a body in motion. They are based on the three Newton’s laws of motion.