MCQ
A particle is projected from ground in vertical upward direction at $t = 0$, with initial velocity $48\, m/s$ and the distance travelled by particle in $5^{th}$ second is......$m$ $(g = 10\, m/s^2)$
- A$3$
- B$115$
- ✓$3.4$
- D$2.8$
✓
Answer
Correct option: C.
$3.4$
c
$5^{\text {th }}$ sec $\Rightarrow t=4 \rightarrow t=5$
$5^{\text {th }}$ sec $\Rightarrow t=4 \rightarrow t=5$
$\mathrm{S}_{4 \rightarrow 8}=$ Displacement in last $0.8\, sec$ of upward journey
$\mathrm{S}=\mathrm{vt}-\frac{1}{2} \mathrm{a}_{\mathrm{y}} \mathrm{t}^{2}, \mathrm{v}=0$
Assuming upward direction to be positive $(+i v e)$
$a_y=-g$
$S_{4 \rightarrow 4.8}=\frac{1}{2} \times g \times(0.8)^{2}=5 \times 0.64$
$\begin{aligned} \mathrm{S}_{4.8 \rightarrow 5}=& \mathrm{ut}+\frac{1}{2} \mathrm{at}^{2} \\ &=0+\frac{1}{2} \times 10 \times(0.2)^{2}=5 \times 0.04 \end{aligned}$
$\mathrm{S}_{4 \rightarrow 5}=5 \times 0.68=\frac{17}{5} \mathrm{m}$
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