A stone falls from a cliff and travels 73.5 m in the last second before it reaches the ground at the foot of the cliff. Find the height of the cliff.
(a) 148.2 m
(b) 184.2 m
(c) 313.6 m
(d) 44.1 m
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A stone falls from a cliff and travels 73.5 m in the last second before it reaches the ground at the foot of the cliff. Find the height of the cliff.
(a) 148.2 m
(b) 184.2 m
(c) 313.6 m
(d) 44.1 m
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Solution :
The stone travels a distance of 73.5 m(S) in the last second( t = 1 s )
Let the initial velocity of the stone at that point be v
Now , as the stone is falling uniformly under the influence of gravity let's apply the second equation of motion to find v
→ S = ut + at² / 2
→ S = v t + at² / 2
→ 73.5 = v x 1 + 9.8 x 1 / 2
→ 73.5 = v + 4.9
→ v = 73.5 - 4.9
→ v = 68.6 m/s
Now let us find the the remaining distance covered by the stone .
By using the third equation of motion we get ,
→ v² = u² + 2as
On rearranging ,
→ S' = v² - u² / 2a
→ S' = 68.6 x 68.6 / 2 x 9.8
→ S' = 4705.96 / 19.6
→ S' = 240 m
Height of the cliff ,
→ H = S + S'
→ H = 240.1 + 73.5
→ H = 313.6 m
The height of the cliff is 313 . 6 m