Question

A man in a balloon rising vertically with an acceleration of \[4.9\,m/{s^2}\] releases a ball 2 sec after the balloon is let go from the ground. The greatest height above the ground reaches by the ball is $(g = 9.8\,m/{s^2})$
A) $14.7\,m$
B) $19.6\,m$
C) $9.8\,m$
D) $24.5\,m$

Answer 1

Hint: In this solution, we will use the second and the third equation of kinematics. We will first find the height of the balloon at 2 seconds and then the maximum possible height that can be achieved by the ball.
Formula used: In this solution, we will use the following formulae:
1- First equation of kinematics: $v = u + at$ where $v$ is the final velocity of an object with initial velocity $u$ under acceleration $a$ in time $t$
2- Second equation of kinematics: $d = ut + \dfrac{1}{2}a{t^2}$where $d$ is the distance travelled by the object with initial velocity $u$ under acceleration $a$ in time $t$
3-Third equation of kinematics: ${v^2} = {u^2} + 2ad$ where $v$ is the final velocity of the object, $d$ is the distance the object travels

Complete step by step answer:
We know that the balloon starts rising with an acceleration of $4.9\,m/{s^2}$ . Initially, the velocity of the balloon is zero. Then in two seconds, it will achieve a height of
$d = 0(2) + \dfrac{1}{2}(4.9){(2)^2}$
$ \Rightarrow d = 9.8\,m$
At this point, the man releases the ball with an initial velocity equal to the velocity of the balloon.
The velocity of the balloon at 2 seconds will be
$v = 0 + (4.9)(2)$
$ \Rightarrow v = 9.8\,m/s$
This will also be the velocity of the ball when it is released. Then the extra height achieved by the ball will be determined from the third equation of kinematics as
${0^2} = {2^2} - 2(9.8)(h)$
Which give us the height
$h = \dfrac{4}{{2 \times 9.8}}$
$ \Rightarrow h = 4.9\,m$
Hence the maximum height of the balloon above the ground will be
$h' = h + d$
$ \Rightarrow h' = 4.9 + 9.8$
Which gives us

$h' = 14.7\,m$ hence the correct choice will be option (A).

Note: When the ball is released from the balloon, the only force that will act on it is gravitational force and hence it will experience gravitational acceleration. The gravitational acceleration will be downwards, which is why we use a negative sign in the third equation of kinematics. After reaching the topmost point, the ball will start falling back to the ground.