Question

A packet dropped from a balloon which is going upwards with the velocity 12m/s, the velocity of the packet after 2 seconds will be:​

Answer 1

Question :–

A packet dropped from a balloon which is going upwards with the velocity 12m/s, the velocity of the packet after 2 seconds will be:

Given :–

1. u = 12m/s
2. t = 2sec
3. g = – 9.8ms²

To Find :–

• The velocity of the packet after 2 seconds will be ?

Solution :–

We know that,

1st equation of motion:-

$\implies v = u + gt$

$\implies v = 12 + (-9.8) \times 2$

$\implies v = 12 + 9.8 \times 2$

$\implies v = 12 - 19.6$

$\implies \boxed{v = - 7.6m/s}$

Answer 2

$\large{\red{\underline{\tt{Given\::-}}}}$

$\green\dashrightarrow\green{\tt u = 12 \:ms^{-1}} \\ \\\green\dashrightarrow \green{\tt t = 2\: sec } \\ \\\green\dashrightarrow\green{\tt g = - 9.8 \: ms^{2}}$

$\large{\red{\underline{\tt{To\:Find:-}}}}$

• The velocity of the packet after 2 seconds will be.

$\large{\red{\underline{\tt{Figure\::-}}}}$

$\setlength{\unitlength}{1.5cm}\begin{picture}(7,2)\thicklines\put(2,0.4){\vector(0,4){1.5}}\put(2,0.2){\circle*{2}}\put(3.3,1.7){\vector(0,-2){1.5}}\put(2.1,1.4){\sf{u = 12 m/s}}\put(1,0){\bf{Packet}}\put(3.4,0.8){\sf{g = 9.8 m/s^\text2 }}\end{picture}$

$\large{\red{\underline{\tt{Solution\::-}}}}$

By using first equation of motion we get :-

$\implies\blue{\tt v = u + gt} \\ \\ \implies\blue {\tt v = 12 - 9.8 \times 2 } \\ \\\implies\blue{\tt v = 12 - 19.6 } \\ \\ \implies\blue{\tt v = -7.6\:ms^{-1}}$

Therefore, the velocity of the packet after 2 seconds will be -7.6 m/s.