Question

A balloon is ascending at the rate of $14\,m/s$ at a height of $98\,m$ above the ground where a packet is dropped from the balloon. After how much time and with velocity does it reach the ground?

Answer 1

Hint: In this problem, we will use Equation of Motion,
$
  v = u + at \\
  s = ut + \dfrac{1}{2}a{t^2} \\
$
where, $s = $ distance/displacement
                  $u = $ initial velocity
                  $a = $ acceleration due to gravity
                  $t = $ time taken
Use the given hint to solve the above problem.

Complete Step-by-Step solution:
Initial velocity= $ + 14\,m/s$(upward)
Acceleration due to gravity= $ - 10\,m/s$(downward)
Displacement= $ - 98\,m$(downward)
Assume, $g = 10\,m/{s^2}$
Now, $s = ut + \dfrac{1}{2}g{t^2}$
$ - 98 = 14 \times t + \dfrac{1}{2}( - 10){t^2}$
$14t - 5{t^2} + 98 = 0$
$5{t^2} - 14t - 98 = 0$
$t = \dfrac{{ - ( - 14) \pm \sqrt {{{\left( {14} \right)}^2} - 4\left( 5 \right)\left( { - 98} \right)} }}{{2 \times 5}}$
$t = \dfrac{{ + 14 \pm \left( {46.43} \right)}}{{10}}$
$t = 6.043\,\sec $
Velocity when it reaches ground:
$v = u + gt$
$v = \left( {14} \right) + \left( { - 10} \right) \times 6.043$
$v = - 46.43\,m/s$
So, time taken , $t = 6.043\,\sec $ and with the velocity ,$v = - 46.43\,m/s$ packet reaches the ground.

Note: In this problem, the sign conventions must be known that when the velocity is directed upwards, a positive $\left( + \right)$ sign is taken. When the velocity is directed downwards, a negative $\left( - \right)$ sign is taken.