Question

A stone is dropped from the top of a cliff. It seems to hit the growth after \[4.2\sec \]. How high is the cliff?

Answer 1

Hint: In kinematics every motion of a body can be expressed in the form of some equations. These equations are called equations of motion. A motion can consist of a body travelling linearly on a ground or a body falling down words from a particular height. We can also have some parabolic motion while hiking a football on the ground. All these situations can be handled by using the equation of motion. Linearly we have three equations of motions.
1. \[v=u+at\]
2. \[{{v}^{2}}-{{u}^{2}}=2as\]
3. \[s=ut+\dfrac{1}{2}a{{t}^{2}}\]
Here v is final velocity
u = initial velocity
a = acceleration of the body
s = displacement covered by the body

Complete answer:
Let the height of the cliff is h.
As we know the initial velocity of the body
\[u=0\]
Using equation 2
\[\begin{align}
  & {{v}^{2}}-{{u}^{2}}=2\times 9.8\times h \\
 & {{v}^{2}}=19.6h\text{ }\left[ a=9.8m/{{s}^{2}}\text{ acceleration due to gravity} \right] \\
 & v=\sqrt{19.6h}\text{ }\left( 4 \right) \\
\end{align}\]
Using third equation of motion
\[\begin{align}
  & h=ut+\dfrac{1}{2}a{{t}^{2}} \\
 & h=0+\dfrac{1}{2}\times 9.8\times {{\left( 4.2 \right)}^{2}} \\
 & h=\dfrac{9.8\times \left( 17.64 \right)}{2} \\
 & h=86.44m\text{ }\left( 5 \right) \\
\end{align}\]
Putting in 4
\[\begin{align}
  & v=\sqrt{19.6\times 86.4} \\
 & v=41.15m/s \\
\end{align}\]

Note:
We take a = acceleration due to gravity as a constant \[=9.8m/{{s}^{2}}\]. It is a uniform acceleration that we attain when a body is falling freely under gravitational effect.
If the body is dropping downwards we use 'a' as positive and if body is thrown upwards we use ‘a' to be negative.
Negative sign indicates that the acceleration is in the opposite direction of earth's gravitational pull.
Acceleration due to gravity ‘a’ doesn't depend upon the mass of the object. If I throw a body of mass 10kg and a body of mass 100 kg (both having the same dimensions) they reach the ground after the same time thrown from the same heights.