Question

A stone is dropped from the top of a cliff and is found to travel 44.1m diving the last sec before it reaches the ground. What is the height of the cliff?

Answer 1

Need to FinD :-

• The height of the cliff .

$\red{\frak{Given}}\Bigg\{ \textsf{The stone travells 44.1 m in last second. }$

Now here , since 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 ,

$\sf\longrightarrow S_n = u + \dfrac{1}{2}a( 2n - 1 ) \\\\\\\sf\longrightarrow 44.1 m = 0m/s +\dfrac{ 1}{2}* 9.8m/s^2 ( 2n -1) \\\\\\\sf\longrightarrow 44.1 m = 4.9 m/s^2 ( 2n - 1 ) \\\\\\\sf\longrightarrow \dfrac{44.1}{4.9} = 2n - 1 \\\\\\\sf\longrightarrow 9 + 1 = 2n \\\\\\\sf\longrightarrow 2n = 10s \\\\\\\sf\longrightarrow \underline{\underline{ n = 5s}}$

• Therefore the stone took 5 second to fall from the Cliff .

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

$\\\sf\longrightarrow s = ut + \dfrac{1}{2}at^2 \\\\\\\sf\longrightarrow s = 0 + \dfrac{ 1}{2} * 9.8m/s^2 * (5s)^2 \\\\\\\sf\longrightarrow s = 4.9 m/s^2 * 25s^2 \\\\\\\sf\longrightarrow \underline{\underline{ \red{ Height_{(cliff)}= 122.5 \ m }}}$