Question

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

Answer 1

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 .

• The initial velocity (u) = 0 m/s [ free fall ]

• Final velocity (v) = 68.6 m/s

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