Question

A body is projected vertically upward with speed $40$ m/s. The distance travelled by body in the last second of upward journey is [ take $g=9.8m/s^2$ and neglect effect of air resistance]

• A. $4.9m$
• B. $9.8m$
• C. $12.4m$
• D. $19.6m$

Answer 1

The correct option is A $4.9m$

$\text{Step 1: Calculate maximum height(H)}$ $\text{[Ref. Fig. 1]}$

(Taking upward direction as positive)

Let $H$ be the maximum height

$V=$ Velocity at top; $V=0m/s$; $a=-9.8m/s^2$; $s=H$; $u=40m/s$

Since acceleration is constant, therefore we can apply equation of motion

$V^2-u^2=2as$

$\Rightarrow 0^2-{40}^2=2(-9.80)H$

$\Rightarrow H=81.63m$

$\text{Step 2: Calculate time required to reach H}$

Applying equation of motion ,

$s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow$ $81.63=40t-\frac{1}{2}\left(9.80\right)t^2$

$\Rightarrow$ $t=4.08s$

$\text{Step 3: Calculate distance travelled by body in last second of upward journey}$ $\text{[Ref. Fig. 2]}$

Let the distance covered in last second $=d$

$\therefore (H-d)m$ distance will be covered in $(t-1)$ seconds

Applying equation of motion,

$s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow (H-d)=40(4.08-1)-\frac{1}{2}\times 9.80(4.08-1{)}^2$

$\Rightarrow$ $81.63-d=76.71$

$\Rightarrow$ $d=4.91m$

Hence option $A$ is correct.