Physics, asked by priyanshucmcl, 9 months ago

A ballon is rising up at 30m/sec .and a stone is released from it when it is at a height 80m above ground . find the time after which stone will hit the ground.

Answers

Answered by amitkumar44481
10

Question :

A balloon is rising up at 30m/s and A stone is released from it when it is at a height 80m above ground. Find the time after which stone will hit the ground.

AnsWer :

After 8 second.

Given value :

 \:  \:  \tt \vec{a}  =  - g \:   \:  \:  \:  \:  \:  \:  \leadsto - 10. \\ \:  \:  \tt  \vec{u} = 30 \: s. \\  \:  \:  \tt  \vec{s} = 80  \: m.

Formula Use :

 \tt  \:  \: \vec{s} = ut +  \frac{1}{2} a {t}^{2}

Solution :

 \tt \leadsto - 80 = 30t +  \frac{1}{2} ( - 10) \times  {t}^{2} .

 \tt \leadsto - 80 = 30t  -  5 {t}^{2} .

 \tt \leadsto - 80 =  -  5 {t}^{2} + 30t .

 \tt \leadsto 0 = -  5 {t}^{2} + 30t + 80 .

 \tt \leadsto 0 = 5 {t}^{2}  - 30t - 80.

  \tt\leadsto0 =  {t}^{2}  - 6t - 16.

  \tt\leadsto0 =  {t}^{2}  - 8t + 2t - 16.

  \tt\leadsto0 =  t(t - 8) + 2(t - 8).

  \tt\leadsto0 =  (t - 8)(t + 2).

  \tt\leadsto0 =  (t + 2)  \\   \leadsto\tt t =  - 2.

And,

  \tt\leadsto0 =  (t - 8) \\  \tt \leadsto t = 8.

Therefore,after 8 second stone hit the ground.

Answered by amitkumar44481
13

Question :

A balloon is rising up at 30m/s and A stone is released from it when it is at a height 80m above ground. Find the time after which stone will hit the ground.

AnsWer :

After 8 second.

Given value :

 \:  \:  \tt \vec{a}  =  - g \:   \:  \:  \:  \:  \:  \:  \leadsto - 10. \\ \:  \:  \tt  \vec{u} = 30 \: s. \\  \:  \:  \tt  \vec{s} = 80  \: m.

Formula Use :

 \tt  \:  \: \vec{s} = ut +  \frac{1}{2} a {t}^{2}

Solution :

 \tt \leadsto - 80 = 30t +  \frac{1}{2} ( - 10) \times  {t}^{2} .

 \tt \leadsto - 80 = 30t  -  5 {t}^{2} .

 \tt \leadsto - 80 =  -  5 {t}^{2} + 30t .

 \tt \leadsto 0 = -  5 {t}^{2} + 30t + 80 .

 \tt \leadsto 0 = 5 {t}^{2}  - 30t - 80.

  \tt\leadsto0 =  {t}^{2}  - 6t - 16.

  \tt\leadsto0 =  {t}^{2}  - 8t + 2t - 16.

  \tt\leadsto0 =  t(t - 8) + 2(t - 8).

  \tt\leadsto0 =  (t - 8)(t + 2).

  \tt\leadsto0 =  (t + 2)  \\   \leadsto\tt t =  - 2.

And,

  \tt\leadsto0 =  (t - 8) \\  \tt \leadsto t = 8.

Therefore,after 8 second stone hit the ground.

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