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Question: A balloon is going upward with velocity \[12\,{\text{m/s}}\]. It releases a packet when it is at a h...

A balloon is going upward with velocity 12m/s12\,{\text{m/s}}. It releases a packet when it is at a height of 65m65\,{\text{m}} from the ground. How much time will the packet take to reach the ground if g=10m/s2g = 10\,{\text{m/}}{{\text{s}}^2}.
A. 5 second
B. 6 second
C. 7 second
D. 8 second

Explanation

Solution

Use the kinematic equation for the motion of the particle. This kinematic equation must give the relation between displacement, initial velocity, acceleration and time. Determine all the physical quantities given in the question and substitute in this kinematic equation to determine the time required for the packet to reach the ground.

Formula used:
The kinematic equation relating the displacement ss, initial velocity uu, acceleration aa and time tt is as follows,
s=ut+12at2s = ut + \dfrac{1}{2}a{t^2} …… (1)

Complete step by step answer:
We have given that the velocity of the balloon is 12m/s12\,{\text{m/s}} and the balloon releases a packet when it is at a height of 65m65\,{\text{m}} from the ground.
We have asked to determine the time required for the packet to reach the ground.
Since the packet is initially in the balloon, it has the same initial velocity as the balloon but with negative sign as the packet will be moving in the downward direction during its travel towards the ground.
u=12m/su = - 12\,{\text{m/s}}
The height of the balloon from the ground is 65m65\,{\text{m}}. Hence, the displacement of the balloon will also be 65m65\,{\text{m}}.
s=65ms = 65\,{\text{m}}
Rewrite equation (1) for the displacement of the packet.
s=ut+12gt2s = ut + \dfrac{1}{2}g{t^2}
Substitute 65m65\,{\text{m}} for ss, 12m/s - 12\,{\text{m/s}} for uu and 10m/s210\,{\text{m/}}{{\text{s}}^2} for gg in the above equation.
(65m)=(12m/s)t+12(10m/s2)t2\left( {65\,{\text{m}}} \right) = \left( { - 12\,{\text{m/s}}} \right)t + \dfrac{1}{2}\left( {10\,{\text{m/}}{{\text{s}}^2}} \right){t^2}
65=12t+5t2\Rightarrow 65 = - 12t + 5{t^2}
5t212t65=0\Rightarrow 5{t^2} - 12t - 65 = 0
5t2+13t25t65=0\Rightarrow 5{t^2} + 13t - 25t - 65 = 0
t(5t+13)5(5t+13)=0\Rightarrow t\left( {5t + 13} \right) - 5\left( {5t + 13} \right) = 0
(5t+13)(t5)=0\Rightarrow \left( {5t + 13} \right)\left( {t - 5} \right) = 0
(t=135) or (t=5)\therefore \left( {t = - \dfrac{{13}}{5}} \right){\text{ or }}\left( {t = 5} \right)

The time cannot be negative. Therefore, the time required for the packet to reach the ground is 5 seconds.Hence, the correct option is A.

Note: One can also solve the same question by another method. One can determine first the distance and time required for the packet to move in an upward direction where its velocity becomes zero. Then determine the time required to reach the ground from this height and finally take the sum of these two times to calculate the final time required.