Question

A particle is dropped from the top of a tower. The distance travelled by it in the last one second is equal to that covered by it in the first three seconds. Find the height of the tower.

Answer 1

Hint: This sum 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 $s=u+\dfrac{a}{2}\left( 2n-1 \right)$.
Formula used: $s=ut+\dfrac{1}{2}a{{t}^{2}}$, $s=u+\dfrac{a}{2}\left( 2n-1 \right)$

Complete step by step solution:
Now, it is given that the particle is dropped from the top of the tower and it covers some distance s in three seconds, so, the distance travelled in given by using the formula,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$ ………………………….(i)
Where, u is initial velocity, a is acceleration, t is time and s is distance travelled.
Now, as the stone is dropped freely it comes down due to gravity of earth so the initial velocity if the stone will be zero and a is equal to gravitational acceleration.
So, $u=0$
$t=3s$
$g=10\ {m}/{{{s}^{2}}}\;$
(for simplicity we will consider the value of ‘g’ as 10 instead of 9.81.)
Substituting the values in expression (i) we will get,
$s=\dfrac{1}{2}\times 10\times {{\left( 3 \right)}^{2}}$
$\Rightarrow s=\dfrac{1}{2}\times 10\times 9=45\ m$ …………………..(ii)
Now, we know that distance travelled by particle in n seconds is given by,
$s=u+\dfrac{a}{2}\left( 2n-1 \right)$ ………………….(iii)
Where, u is initial velocity, a is acceleration and time at $n^{th}$ second, and s is distance travelled.
Now, here also initial velocity is zero and a is equal to gravitational acceleration.
So, the travelled by the particle in $n^{th}$ second before striking the ground is given by,
$s=\dfrac{g}{2}\left( 2n-1 \right)$
$\Rightarrow s=\dfrac{10}{2}\left( 2n-1 \right)$
$\Rightarrow s=10n-5$ …………………..(iv),
Now, equating the value of expression (ii) with expression (iv) we get,
$10n-5=45\Rightarrow 10n=45+5=50$
$\Rightarrow 10n=50\Rightarrow n=\dfrac{50}{10}=5$
Now, the height can be derived by substituting the value of n in equation (i) which is given by,
$h=\dfrac{1}{2}g{{t}^{2}}$
$\Rightarrow h=\dfrac{1}{2}\left( 10 \right){{\left( 5 \right)}^{2}}$
$\Rightarrow h=5\times 25=125\ m$
So, the height of the tower is 125 m.

Note: In such questions, students might forget to consider that during free fall the initial velocity is zero so students must take care while solving such problems. Students might also forget to use the value of n in place of t to find the final answer and due to that also they might face problems so they should take care of this, while solving such problems.