Question

A particle is dropped from the top of a tower. It covers $40m$ in the last $2s$. Find the height of the tower.

Answer 1

Hint: Use kinematics equation to calculate the velocity at the starting of last $2s$.From this value of velocity calculate the distance covered in the first half by the kinematics formula. Total height of the tower will be the sum of both the heights.

Formula used:
 $s=ut+\dfrac{1}{2}a{{t}^{2}}$
Where, $s=\text{displacement, }u=\text{ initial velocity,}a=\text{ acceleration}$
If particle falls under influence of gravity then $a=g=\text{ acceleration due to gravity}$
${{v}^{2}}-{{u}^{2}}=2as$
Where ,
$\begin{align}
  & v=\text{ final velocity of the particle} \\
 & u\text{ =initial velocity of particle} \\
 & s\text{ =displacement} \\
 & a\text{=acceleration} \\
\end{align}$

Complete step by step answer:
Suppose the height of the tower is $H$. When the particle is at height $h=H-40$, let its velocity be . So when the particle is at height $'h'$ from the top let its velocity be $'u'$. According to the question the particle covers the last $40m$ in $2\sec $.
$s=ut+\dfrac{1}{2}g{{t}^{2}}$
Here $s=40m$,$t=2\sec $,$g=\text{ acceleration due to gravity=}10m{{s}^{-2}}$
Putting these values we get.

$\begin{align}
  & 40=u\times 2+\dfrac{1}{2}\times 10\times {{2}^{2}} \\
 & \Rightarrow 40=2u+20 \\
 & \Rightarrow 2u=20 \\
 & \Rightarrow u=10m{{s}^{-1}} \\
\end{align}$

i.e. when the particle starts to cover the last $40m$ distance its velocity is $10m{{s}^{-1}}$.
When the object dropped from the top of the tower its initial velocity is $u'=0$. After covering the distance $'h'$ its final velocity will be the initial velocity of the object when it starts to cover the last $40m$ which is $v=10m{{s}^{-1}}$.
According to formula ${{v}^{2}}-{{u}^{2}}=2as$
Here $v=10m{{s}^{-1}},u=u'=0,a=g=10m{{s}^{-2}}\text{ and }s=h$
Now
$\begin{align}
  & {{10}^{2}}-{{0}^{2}}=2\times 10\times h \\
 & \Rightarrow 20h=100 \\
 & \Rightarrow h=5m \\
\end{align}$
Total height of tower is $H+h=40m+5m=45m$

Additional Information
If a particle has initial velocity $u$ have constant acceleration $a$in time $t$ then the final velocity is given by $v=u+at$

Note:
All the equation of kinematics $v=u+at$,${{v}^{2}}-{{u}^{2}}=2as$,and $s=ut+\dfrac{1}{2}g{{t}^{2}}$ can be used only for uniformly accelerated motion. It is not valid for the motion where the acceleration changes with time. Also if a particle falls under the influence of gravity(free fall) the acceleration of the object is constant and is equal to acceleration due to gravity. If an object is thrown upward then $a=-g$ and of falls then $a=g$.