Physics, asked by marshal93, 1 year ago

a stone is thrown vertically upward with a speed of 40 metre per second the time interval for which particle of about 40 m from the ground is dash​

Answers

Answered by BrainlyConqueror0901
11

{\bold{\underline{\underline{Answer:}}}}

{\bold{\therefore Time\:taken=6.8\:sec}}

{\bold{\underline{\underline{Step-by-step\:explanation:}}}}

• In the given question information given about a stone is thrown vertically upward with a speed of 40 m/s.

• We have to find the time interval at a height of 40 m from the ground.

 \underline \bold{Given : } \\  \implies Initial \: velocity(u) = 40 \: m/s \\  \\ \implies Distance (s)= 40 \: m \\  \\  \implies Acceleration(a) =  - 10 \: m /{s}^{2}  \\  \\  \underline \bold{To \: Find : } \\  \implies Time \:taken(t)= ?

• According to given question :

  \bold{Using \: second \: equation \: of \: motion :} \\   \implies  {v}^{2}  =  {u}^{2}   + 2as \\  \\  \implies  {v}^{2}  =   {40}^{2}  + 2 \times ( - 10) \times 40 \\  \\  \implies  {v}^{2}  =1600 - 800 \\  \\  \implies  {v}^{2}  = 800 \\  \\  \implies v =  \sqrt{800}  \\  \\   \bold{\implies v = 20 \sqrt{2}   \: m/s}\\  \\   \bold{Using \: first \: equation \: of \: motion : } \\  \implies v = u + at \\  \\  \implies 20 \sqrt{2}  = 40 + ( - 10) \times t \\  \\  \implies 20 \sqrt{2}  - 40 =  - 10t \\  \\  \implies t =  \frac{20  \sqrt{2} -40 }{ - 10}  \\  \\  \implies t =  \frac{ - 10(2 \sqrt{2} + 4) }{ - 10}  \\  \\  \implies t = 2 \times 1.4 + 4 \\  \\   \bold{\implies t = 6.8 \: sec}

Answered by jatindevrajput
9

     \implies  {v}^{2}  =  {u}^{2}   + 2as \\  \\  \implies  {v}^{2}  =   {40}^{2}  + 2 \times ( - 10) \times 40 \\  \\  \implies  {v}^{2}  =1600 - 800 \\  \\  \implies  {v}^{2}  = 800 \\  \\  \implies v =  \sqrt{800}  \\  \\   \bold{\implies v = 20 \sqrt{2}   \: ms}\\   \\  \implies v = u + at \\  \\  \implies 20 \sqrt{2}  = 40 + ( - 10) \times t \\  \\  \implies 20 \sqrt{2}  - 40 =  - 10t \\  \\  \implies t =  \frac{20  \sqrt{2} -40 }{ - 10}  \\  \\  \implies t =  \frac{ - 10(2 \sqrt{2} + 4) }{ - 10}  \\  \\  \implies t = 2 \times 1.4 + 4 \\  \\   \bold{\implies t = 6.8 \: sec}

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