Physics, asked by krsaivarun, 7 days ago

A body fall freely from the top of a cliff and during the last second, it falls 11/36 of the whole height. What is the height of the cliff? ( g = 9.8m/sec2)

Answers

Answered by Anonymous
6

Given:-

  • Initial Velocity of body = 0m/s

  • Acceleration due to gravity = +9.8m/s²

  • It falls 11/36 of the whole Height in it's last second.

To Find :-

  • Height of the cliff.

Formulae used:-

  • S = ut + ½ × a × t²

Where,

  • s = Height
  • u = Initial Velocity
  • a = Acceleration
  • t = Time

How To Solve

  • Here, we don't know Height so we will say it as "x"

  • It is also given that it falls 11/36 of the height at last second so we will take t = (t - 1)

Now,

→ s = ut + ½ × a × t

→ x = 0 + ½ × g × t

→ x = ½gt .....................1

Now, Atq

\sf{ s = \dfrac{11x}{36}}

\sf{ t = ( t - 1 ) }

Therefore,

\sf{ s = ut + \dfrac{1}{2}\times{a}\times{(t)^2}}

\sf{ \dfrac{11x}{36} = 0 + \dfrac{1}{2}\times{a}\times{ (t - 1)^2}}.......2

Putting the value of eq 1 in eq 2

\sf{ \dfrac{11}{36}\times{\dfrac{1}{2}gt^2} = \dfrac{1}{2}\times{g}\times{ (t - 1)^2}}

\sf{ \dfrac{11t^2}{36} = ( t - 1)^2 }

\sf{ 11t^2 = 36 ( t^2 - 2t + 1)}

\sf{ 11t^2 = 36t^2 - 72t + 36}

\sf{ 25t^2 - 72t + 36 = 0}

→ From above Equation we get t = 6 second.

Now, Putting the Value of t in eq 1

→ x = ½ × gt²

→ x = ½ × 9.8 × (6)²

→ x = ½ × 9.8 × 36.

→ x = 9.8 × 18

→ x = 176.4.

Hence, The Height of the cliff is 176.4m.

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