Question

A bullet is fired vertically upward with an initial velocity of $50{\text{ms}}^{-1}$. If $g=10{\text{ms}}^{-2}$, what is the ratio of the distances travelled by the bullet during the first and last second of its upward motion ?

• A. $9:1$
• B. $9:2$
• C. $3:1$
• D. $9:4$

Answer 1

The correct option is A $9:1$

As we know that the time taken to reach the maximum height is
$t=\frac{u}{g}=\frac{50}{10}=5\text{sec}$
Also, displacement in $n^{th}$ sec is given as
$S_{n^{th}}=u+\frac{a}{2}[2n-1]$
As per the question we have
$\frac{S_1}{S_{5^{th}}}=\frac{50-\frac{10}{2}\left[2\right(1)-1]}{50-\frac{10}{2}\left[2\right(5)-1]}$
$\Rightarrow \frac{S_1}{S_{5^{th}}}=\frac{50-5}{50-45}=\frac{45}{5}$
$\Rightarrow \frac{S_1}{S_{5^{th}}}=\frac{9}{1}$
Hence, the ratio is $9:1$