Question

A body is thrown vertically up with velocity ‘u’. The distance travelled by it in \[{5^{th}}\] and \[{6^{th}}\] seconds are equal. The velocity u is given by $\left( {g = 9.8\dfrac{m}{{{s^2}}}} \right)$
(A) $24.5\dfrac{m}{s}$
(B) $49.0\dfrac{m}{s}$
(C) $73.5\dfrac{m}{s}$
(D) $98.0\dfrac{m}{s}$

Answer 1

Hint: If a body travels an equal distance at a certain time span then it is possible only when the body reaches maximum height at one of time intervals and after that body falls freely under gravity. We make use of the equation of motion to find the initial velocity of the particle.

Complete step by step answer:
It is given that the distance travelled in \[{5^{th}}\] and \[{6^{th}}\] seconds are equal, it is possible only when the body reaches maximum height at 5th second and after that body falls freely under gravity.
Then, the distance travelled when t = 5 sec is given by:
We know that, distance travelled in nth second is,
\[{S_n} = u + \dfrac{a}{2}\left( {2n - 1} \right)\]
Where,
u = Initial velocity
a = g = Acceleration of the body
Substituting the values in above equation we get,
\[ \Rightarrow {S_5} = u + \dfrac{g}{2}\left( {2\left( 5 \right) - 1} \right)\]
\[ \Rightarrow {S_5} = u - \dfrac{{9g}}{2}\] … (i)
$a = - g$, because the body is moving vertical upward direction.
Similarly, distance travelled at \[{6^{th}}\] second is, i.e., ${S_5}$ in 1 second:
\[{S_6} = ut + \dfrac{1}{2}a{t^2}\]
\[ \Rightarrow {S_6} = 0 + \dfrac{g}{2}\left( {{1^2}} \right) = \dfrac{g}{2}\] … (2)
Since distance travelled at \[{5^{th}}\] and \[{6^{th}}\] seconds are equal, then we have:
${S_5} = {S_6}$
$ \Rightarrow \dfrac{g}{2} = u - \dfrac{{9g}}{2}$
$ \Rightarrow u = \dfrac{g}{2} + \dfrac{{9g}}{2}$
$ \Rightarrow u = \dfrac{{10g}}{2} = \dfrac{{10 \times 9.8}}{2} = 49\dfrac{m}{s}$
Therefore, the body was thrown vertically upwards with $49\dfrac{m}{s}$ .

Hence, option B is correct.

Note:
One should remember that it is not necessary to always solve problems conceptually. By understanding the fact that during \[{5^{th}}\] sec the body is moving upward and at \[{6^{th}}\] second object is coming downwards, hence we can conclude that at t = 5 sec object is at its maximum height and v = 0, then by equation of motion:
$v = u + at$
$ \Rightarrow 0 = u - 5g$
$ \Rightarrow u = 49\dfrac{m}{s}$