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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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__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

→

→

**Theref****ore**,

→

→ ....**...2**

Putting the value of eq 1 in eq 2

→

→

→

→

→

→ From above Equation we get t = 6 second.

**Now****,**** ****Putt****ing**** ****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**** ****Heig****ht**** ****of**** ****the**** ****cliff**** ****is**** ****1****7****6****.****4****m****.**

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