Question

A body is through vertically upwards with an initial velocity '$u$' reaches a maximum height in$6s$. What is the ratio of the distance traveled by the body in the first second to the seventh second?
$(A)\;1:1$
$(B)11:1$
$(C)1:2$
$(D)1:11$

Answer 1

When 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 the time intervals and after that body falls freely under gravity. We make use of the equation of motion to find the ratio of the distance traveled by the body in the first second to the seventh second.

Formula used:
The distance traveled by ${n^{th}}$second is given by
${s_n} = u + \dfrac{a}{2}\left( {2n - 1} \right)$

Complete step-by-step solution:
From the given data, the body reaches the maximum height $6s$. Let $u$be the initial velocity. At maximum height, the velocity is zero. Using the equation of motion as
$v = u + at$
$\Rightarrow 0 = u - (10)(6)$
$\Rightarrow u = 60m/s$
Displacement in the first second is ${S_1} = 60 - \left( {\dfrac{{10}}{2}} \right)\left( {2 - 1} \right) = 55m$
Displacement in the seventh second is ${S_2} = 60 - \left( {\dfrac{{10}}{2}} \right)\left( {14 - 1} \right) = - 5m$
$\dfrac{{{S_1}}}{{{S_2}}} = \dfrac{{55}}{5} = \dfrac{{11}}{1}$
The ratio of the distance traveled by the body in the ${1^{st}}$ second to the ${7^{th}}$ second is$11:1$.
Hence, option B is correct.

Note: We need to know the phenomena of a body thrown vertically upwards under the influence of gravity, its properties, and its trajectory in order to answer this type of question. Having knowledge of this concept helps us directly relate to when a body can have an equal displacement at two different time intervals. While a body is thrown upwards with an initial velocity$\left( u \right)$, it keeps gaining its potential energy$\left( {P.E} \right)$ and losing its kinetic energy$\left( {K.E} \right)$ as it goes upwards and vice versa, when it falls down but the total energy of the body is conserved at all times.