Question

A body is thrown up with a velocity $29.23{\text{ m}}{{\text{s}}^{ - 1}}$ .The distance travelled in last but one second of upward motion is
$A.\,{\text{ 14}}{\text{.7 m}}$
$B.\,{\text{ 9}}{\text{.8 m}}$
$C.\,{\text{ 4}}{\text{.9 m}}$
$D.\,{\text{ 1 m}}$

Answer 1

Hint: in order to solve the question, we will first use the newton’s third law of motion to find the time then we will use the formula of ${n^{th}}$ second displacement formula and then that by using the newton’s second law of motion we will replace the n second displacement and n-1 second displacement after then we just substitute the value of the physical quantity and we will get the answer

Formula used:
Newton’s third law of motion
$v = u + at$
Here, $v$ refers to final velocity, $u$ refers to initial velocity, $a$ refers to acceleration and $t$ refers to time period.
Newton’s second law of motion
$s = ut + \dfrac{1}{2}a{t^2}$
Here, $s$ refers to displacement, $u$ refers to initial velocity, $a$ refers to acceleration and $t$ refers to time period.
Displacement in ${n^{th}}$ second
${S_{{n^{th}}}} = {S_n} - {S_{n - 1}}$
Here, ${S_n}$ refers to displacement in n second and ${S_{n - 1}}$ refers to displacement in (n-1) second.

Complete step by step answer:
In the question we are asked to find time to cover upward motion.(i.e. last second ) for which they have given initial velocity,\[u{\text{ }} = \;{\text{ }}29.93{\text{ }}m{s^{ - 1}}\]. At max height, final velocity, v =0. Acceleration due to gravity, $a = - 9.8{\text{ }}m{s^{ - 2}}$.By using the Newton' s third equation of motion we will find the time
$v = u + at$
Now we will insert the value of initial velocity, final velocity and acceleration
$0 = 29.23\,m{s^{ - 1}} - 9.8 \times t$
$\Rightarrow t = 2.98{\text{ }}s$
Hence the time came to be $t = 2.98{\text{ }}s$.

Now by using the formula of displace we will find the distance covered in ${n^{th}}$
\[\;Displacement{\text{ }}in\;second\;\; = {\text{ }}Displacement{\text{ }}in{\text{ }}'n'{\text{ }}seconds\; - {\text{ }}Displacement{\text{ }}in{\text{ }}\left( {n - 1} \right){\text{ }}seconds\;\]
${S_{{n^{th}}}} = {S_n} - {S_{n - 1}}$ ……………(1)
Now we will use newton's second law of motion to substitute
${S_n}{\text{ and }}{S_{n - 1}}$
$\Rightarrow {S_n} = un + \dfrac{1}{2}a{n^2}$ ……….….(2)
$\Rightarrow {S_{n - 1}} = u(n - 1) + \dfrac{1}{2}a{(n - 1)^2}$ ……….….(3)

Substituting the value of ${S_n}{\text{ and }}{S_{n - 1}}$ in equation 1 using equation 2 and equation 3
${S_{{n^{th}}}} = [un + \dfrac{1}{2}a{n^2}] - [u(n - 1) + \dfrac{1}{2}a{(n - 1)^2}]$
Solving the equation further we get
${S_{{n^{th}}}} = u + \dfrac{1}{2}a(2n - 1)$
Now substituting these values in the above equation we can the displacement
${S_{{n^{th}}}} = u + \dfrac{1}{2}a(2t - 1)$
$ \Rightarrow {S_{{n^{th}}}} =29.23{\text{ }}m{s^{ - 1}} + \dfrac{1}{2}( - 9.8m{s^{ - 2}})(2 \times 2.98 - 1)$
$ \Rightarrow {S_{{n^{th}}}} =29.23 - 24.30$
$ \therefore {S_{{n^{th}}}} =4.9{\text{ m}}$ (approx.)
Hence,displacement in the last second of upward motion is $4.9{\text{ m}}$.

Hence, the correct answer is option C.

Note: Many of the students can make the mistake by using the different time of reference by not taking initial velocity as $29.23{\text{ }}m{s^{ - 1}}$ and final velocity at top to zero any other reference will only make the question difficult and exceptionally in this question we have to assume like this along with this gravity is negative because it is applied downward in question we are moving upward.