Question

A particle is projected vertically upwards with speed $20\text{m/s}$ from top of a tower of height $20\text{m}$ as shown in figure. Given $B$ is top most point of trajectory and $C$ is at same height as $A$ :
$\begin{array}{cccc}& \text{Column - I}& & \text{Column - II}\\ \left(A\right)& \text{ratio of maximum height from ground (BD)}& \left(P\right)& \frac{1}{\sqrt{2}}\\ & \text{to the initial height from ground (AD)is}& & \\ \left(B\right)& \text{ratio of distance travelled in 1}{}^{st}\text{second to}& \left(Q\right)& \frac{1}{\sqrt{3}}\\ & \text{the distance travelled in 2}{}^{nd}\text{second is}& & \\ \left(C\right)& \text{ratio of initial speed at A to the final}& \left(R\right)& 1\\ & \text{just before reaching to ground (D) is}& & \\ \left(D\right)& \text{ratio of time taken from A to C and time}& \left(S\right)& 2\\ & \text{taken from A to B is}& & \\ & & \left(T\right)& 4\\ & & \left(U\right)& 3\end{array}$
Which of the following option has the correct combination considering column-I and column-II

• A. $B\to P$
• B. $A\to S$
• C. $B\to R$
• D. $A\to Q$

Answer 1

The correct option is B $A\to S$

(I) $AB=$ maximum height $=\frac{v^2}{2g}=\frac{20\times 20}{2\times 10}=20\text{m}$
$\therefore \frac{BD}{AD}=\frac{20+20}{20}=2$

(II) $\frac{s\left(1^{st}\right)}{s\left(2^{nd}\right)}=\frac{20\times 1-\frac{1}{2}\times 10\times 1^2}{20-\left[15\right]}=\frac{15}{5}=3$