Question

A stone is dropped from the top of a cliff and is found to travel 14.7 m in the last second before it reaches the ground. Find the height of the cliff.​

Answer 1

Need to FinD :-

• The height of the cliff .

$\red{\frak{Given}}\Bigg\{ \textsf{The stone travells 14.7 m in last second. }$

Now here , send the stone was dropped from the cliff therefore the initial velocity of the stone will be zero. The stone will accelerate due to the acceleration due to gravity , with 9.8 m/s² .

• Now substituting the values in the formula to find the distance travelled in nth second ,

$\sf\longrightarrow S_n = u + \dfrac{1}{2}a( 2n - 1 ) \\\\\\\sf\longrightarrow 14.7 m = 0m/s +\dfrac{ 1}{2}* 9.8m/s^2 ( 2n -1) \\\\\\\sf\longrightarrow 14.7 m = 4.9 m/s^2 ( 2n - 1 ) \\\\\\\sf\longrightarrow \dfrac{14.7}{4.9} = 2n - 1 \\\\\\\sf\longrightarrow 3 + 1 = 2n \\\\\\\sf\longrightarrow 2n = 4s \\\\\\\sf\longrightarrow \underline{\underline{ n = 2s}}$

• Therefore the stone took 2 seconds to fall from the Cliff .

• Now on substituting the values in second equation of motion we have ,

$\\\sf\longrightarrow s = ut + \dfrac{1}{2}at^2 \\\\\\\sf\longrightarrow s = 0 + \dfrac{ 1}{2} * 9.8m/s^2 * (2s)^2 \\\\\\\sf\longrightarrow s = 4.9 m/s^2 * 4s^2 \\\\\\\sf\longrightarrow \underline{\underline{ \red{ Height_{(cliff)}= 19.6 \ m }}}$

Answer 2

Answer:

The height of the cliff .

\red{\frak{Given}}\Bigg\{ \textsf{The stone travells 14.7 m in last second. }Given{The stone travells 14.7 m in last second.

Now here , send the stone was dropped from the cliff therefore the initial velocity of the stone will be zero. The stone will accelerate due to the acceleration due to gravity , with 9.8 m/s² .

Now substituting the values in the formula to find the distance travelled in nth second ,

\begin{gathered}\sf\longrightarrow S_n = u + \dfrac{1}{2}a( 2n - 1 ) \\\\\\\sf\longrightarrow 14.7 m = 0m/s +\dfrac{ 1}{2}* 9.8m/s^2 ( 2n -1) \\\\\\\sf\longrightarrow 14.7 m = 4.9 m/s^2 ( 2n - 1 ) \\\\\\\sf\longrightarrow \dfrac{14.7}{4.9} = 2n - 1 \\\\\\\sf\longrightarrow 3 + 1 = 2n \\\\\\\sf\longrightarrow 2n = 4s \\\\\\\sf\longrightarrow \underline{\underline{ n = 2s}}\end{gathered}

⟶S

n

=u+

2

1

a(2n−1)

⟶14.7m=0m/s+

2

1

∗9.8m/s

2

(2n−1)

⟶14.7m=4.9m/s

2

(2n−1)

⟶

4.9

14.7

=2n−1

⟶3+1=2n

⟶2n=4s

⟶

n=2s

Therefore the stone took 2 seconds to fall from the Cliff .

Now on substituting the values in second equation of motion we have ,

\begin{gathered}\\\sf\longrightarrow s = ut + \dfrac{1}{2}at^2 \\\\\\\sf\longrightarrow s = 0 + \dfrac{ 1}{2} * 9.8m/s^2 * (2s)^2 \\\\\\\sf\longrightarrow s = 4.9 m/s^2 * 4s^2 \\\\\\\sf\longrightarrow \underline{\underline{ \red{ Height_{(cliff)}= 19.6 \ m }}}\end{gathered}

⟶s=ut+

2

1

at

2

⟶s=0+

2

1

∗9.8m/s

2

∗(2s)

2

⟶s=4.9m/s

2

∗4s

2

⟶

Height

(cliff)

=19.6 m