Question

A stone is dropped from the top of a cliff and is found to travel $44.1\ m$ diving at the last second before it reacts to the ground. What is the height of the cliff? $g=9.81\ {m}/{{{s}^{2}}}\;$.

Answer 1

- Hint: This question can be solved using the equations which shows relations between velocity, acceleration, time and distance and they are given in the mathematical form as,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$ and ${{v}^{2}}-{{u}^{2}}=2as$.

Formula used: $s=ut+\dfrac{1}{2}a{{t}^{2}}$, ${{v}^{2}}-{{u}^{2}}=2as$

Complete step-by-step solution -
Now, in the question we are given that distance travelled by the stone before striking the ground in last second so from this we can say that,
$s=44.1\ m$
$t=1s$
And here the acceleration is taken as gravitational acceleration so,
$a=g=9.81\ {m}/{{{s}^{2}}}\;$
Now, using the expression of distance travelled in time t with acceleration a and initial velocity is given by,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$ …………………………(i)
Where, s is distance travelled, t is time, a is acceleration, u is initial velocity.
So, substituting the values in expression (i) we will get,
$44.1=u\left( 1 \right)+\dfrac{1}{2}\left( 9.81 \right){{\left( 1 \right)}^{2}}$
$\Rightarrow 44.1=u+4.905$
$\Rightarrow u=44.1-4.905=39.195\ {m}/{s}\;$
So, initial velocity is $39.195\ {m}/{s}\;$.
Now, we know that relation between final velocity, acceleration and initial velocity is given by,
${{v}^{2}}-{{u}^{2}}=2as$ ……………………(ii)
Where, v is final velocity, u is initial velocity, a is acceleration and s is distance travelled.
Now, here v will be u of previous expression as the initial velocity becomes the final velocity of the stone just before striking the ground so it can be given as,
$u=0\ $
$v=39.195\ {m}/{s}\;$
$a=g=9.81\ {m}/{{{s}^{2}}}\;$
So, using the above values we will find the distance travelled,
${{\left( 39.195 \right)}^{2}}-{{\left( 0 \right)}^{2}}=2\left( 9.81 \right)s$
$\Rightarrow \dfrac{1536.24}{2\times 9.81}=s$
$\Rightarrow s=78.3\ m$
Now, the total distance travelled by the stone before striking the ground is,
$\text{total distance}=78.3+44.1=122.4\ m$
Now, the total distance travelled by the stone is equal to the height of the cliff, so the height of the cliff is $122.4\ m$.

Note: This sum can also be solved by calculating the time of flight and potential energy of the stone and thus the answer can be derived. Students might make mistakes in considering the final velocity of stone before striking ground as initial velocity derived so the answer would be wrong if they do that, so, students should understand the problem and solve it carefully.