Question

A packet is released from a balloon which is moving upward when the balloon is at a height 100m above ground. The packet reaches the ground in 8 seconds. Find the speed of the balloon when the packet is released. Assume ${\text{g = 10m}}{{\text{s}}^{{\text{ - 2}}}}$ .
(A) ${\text{28m}}{{\text{s}}^{{\text{ - 1}}}}$
(B) ${\text{20m}}{{\text{s}}^{{\text{ - 1}}}}$
(C) ${\text{15m}}{{\text{s}}^{{\text{ - 1}}}}$
(D) ${\text{10m}}{{\text{s}}^{{\text{ - 1}}}}$

Answer 1

Hint : Once the packet is released from the balloon, it has uniformly accelerated motion. It falls under the sole influence of gravity. So we need to use the laws of motion and by substituting the known values from the question we will get the answer.

Formula Used: The following formulae are used in solving the problem,
$S = ut + \dfrac{1}{2}a{t^2}$ where $S$ is the distance travelled, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time taken.

Complete step by step answer
Newton’s equations of motion are used to describe the behaviour of an object in a reference frame with respect to its velocity, acceleration and distance as a time function.
An object that falls solely under the influence of gravity is a freely falling body. The motion of a freely falling body is an example of a motion where the acceleration does not change with the function of time. It is thus a uniformly accelerated motion. The equations of motion are expressed in terms of velocity, acceleration and displacement. These are vector quantities that are dependent on direction.
The packet, which falls from the balloon, falls towards the ground under the influence of gravity. The acceleration of the body is equal to the acceleration due to gravity.
According to Newton’s second equation of motion,
$S = ut + \dfrac{1}{2}a{t^2}$ where $S$ is the distance travelled, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time taken.
In case of the packet that is dropped from an upward moving balloon, acceleration $a$ = acceleration due to gravity $g$ . We also have $S = H$ .
Thus, this equation can be rewritten as,
$H = - ut + \dfrac{1}{2}g{t^2}$ , where $H$ is the height above ground from which the freely falling object is dropped.
It has been given in the question that, the height above ground from which the packet is dropped is 100m and the time taken to reach ground is 8 seconds. The initial velocity $u$ of the packet is the same as the velocity of the balloon at the time when the packet was dropped. Since the motion of the balloon is upwards, away from the ground and that of the packet is towards the ground, the direction of velocity is negative.
Therefore, assigning the values to the variables we get,
$H = 100m$ , $t = 8s$ and $g = 10m{s^{ - 2}}$ .
Substituting these in the equation of motion,
$100 = - u \times 8 + \dfrac{1}{2} \times 10 \times {(8)^2}$
$\Rightarrow 100 = - 8u + 5 \times 64$
Taking $8u$ on the other side,
$8u = 320 - 100$
$\Rightarrow 8u = 220$
Thus, we get the value of $u$ ,
$u = \dfrac{{220}}{8}$
$\Rightarrow u = 27.5m{s^{ - 1}} \approx 28m{s^{ - 1}}.$
$\therefore$ The approximate velocity of the balloon is $28m{s^{ - 1}}$ . Thus, the correct option is Option A.

Note
It is to be noted that the directions of velocity, acceleration and displacement are very important as they determine the signs before the values in the equation. For a freely falling body, under the influence of gravity, acceleration due to gravity, acts towards the ground and is thus positive. Velocity is in the opposite direction, and is negative.