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
→
→
Therefore,
→
→ .......2
Putting the value of eq 1 in eq 2
→
→
→
→
→
→ 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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