Question

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)
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Answer 1

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.