Question

A stone falls from a cliff and travels 24.5m in the last second before it reaches the ground at the foot of the cliff. find the height of the cliff.

Answer 1

Here is the answer---

Let the total time taken by the stone to fall from the cliff be x seconds.

Using Second Equations of Motion,
S = ut + 1/2 × gt²
Where,
S = Total Height
u = Initial Velocity
= 0
t = total time
= x
g = acceleration due to gravity.
= 9.8 m/s²

Now,
S = 0 + 1/2 × 9.8x²
= 4.9x² m.

Distance travelled by the stone in last second = 24.5 m.

Distance travelled by the stone in (x - 1) seconds = 0 + 1/2 × 9.8 × (x - 1)²
= 4.9(x - 1)²

We know,
Total Distance travelled by the stone - Distance travelled by the Stone in (x - 1) seconds = Distance travelled by the stone in last seconds.

4.9x² - 4.9(x - 1)² = 24.5
4.9[x² + (x - 1)²] = 24.5
x² + x² + 1 - 2x = 5
2x² - 2x + 1 = 5
2(x² - x) = 4
x² - x = 2
x² - x - 2 = 0
⇒ x² - 2x + x - 2 = 0
⇒ x(x - 2) + 1(x - 2) = 0
⇒ (x - 2)(x + 1) = 0
⇒ x = 2 and x = -1

Since, Time cannot be negative, ∴ Rejecting x = -1 seconds,
Thus, Time = x = 2 seconds.

Now,
S = 4.9x²
= 4.9 × (2)²
= 4.9 × 4
= 19.6 m.


∴ Height of the Cliff is 19.6 m