Question

A ball is dropped from a balloon going up at a speed of $7\,m/s$ . If the balloon was at a height of $60\,m$ at the time of dropping the ball, how long will the ball take in reaching the ground?

Answer 1

Hint:
Here we have to apply the second equation of motion to get the answer.
There are three equations of motion.

Complete step by step solution:
In physics, equations of motion are classified as equations that characterise a physical system’s behaviour in terms of its motion as a function of time. For deriving components such as displacement, velocity, time and acceleration, there are three motion equations that can be used.
The second motion equation gives an object’s displacement under continuous acceleration.
The second equation of motion is mathematically given by-
$s = ut + \dfrac{1}
{2}a{t^2}$

Where, $s$ is the distance,
$u$ is the initial velocity,
$a$ is the acceleration and
$t$ is the time taken.
Given,
A ball is dropped from a balloon going up at a speed of $7\,m/s$.
The balloon was at a height of $60\,m$ at the time of dropping the ball.
Here we take the speed as the initial velocity and the height as the distance.
So,
$u = 7m/s$

Since, the ball was initially going upwards, so,
$u = - 7\,m/s$

$h = 60\,m$

$g = 10m/{s^2}$

We know that,
$s = ut + \dfrac{1}
{2}a{t^2}$

If acceleration is not given in the question then we have to take acceleration due to gravity instead of acceleration.
$s = ut + \dfrac{1}{2}g{t^2}$
Where $g$ is the acceleration due to gravity.
Putting the values in the above equation, we get-
$
  s = ut + \dfrac{1}
{2}g{t^2} \\
  60 = - 7t + \dfrac{1}
{2}10{t^2} \\
  5{t^2} - 7t - 60 = 0 \\
$

Now, applying the formula of quadratic equation we get-
$
  t = \dfrac{{7 \pm \sqrt {49 + 4 \times 5 \times 60} }}
{{2 \times 5}} \\
  t = \dfrac{{7 \pm 35.34}}
{{10}} \\
$

Considering only the positive value we get-
$t = \dfrac{{7 + 35.34}}
{{10}} = 4.2s$

Hence, the ball will take $4.2\,s$ to reach the ground.

Note:
Here we have to be careful while putting the value of initial velocity. We have to consider negative values here since the velocity was initially in the opposite direction of motion otherwise we would get a different answer.