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 $A$.
List-I gives certain statements regarding particle
List - II gives corresponding results in SI units.
Match the statements in List-I with corresponding results in List-II.

$\begin{array}{cccc}& \text{List-I}& & \text{List-II}\\ \text{(I)}& \text{ratio of maximum height from ground (BD)}& \text{(P)}& \frac{1}{\sqrt{2}}\\ & \text{to the initial height from ground (AD) is}& & \\ \text{(II)}& \text{ratio of distance travelled in 1}{}^{st}\text{second to}& \text{(Q)}& 1\\ & \text{the distance travelled in 2}{}^{nd}\text{second is}& & \\ \text{(III)}& \text{ratio of initial speed at A to the final}& \text{(R)}& 2\\ & \text{just before reaching to ground (D) is}& & \\ \text{(IV)}& \text{ratio of time taken from A to C and time}& \text{(S)}& 3\\ & \text{taken from A to B is}& & \end{array}$

• A. $I\to R;II\to S;III\to R;IV\to P$
• B. $I\to S;II\to R;III\to P;IV\to R$
• C. $I\to R;II\to S;III\to P;IV\to R$
• D. $I\to P;II\to S;III\to Q;IV\to R$

Answer 1

The correct option is C $I\to R;II\to S;III\to P;IV\to R$

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

(III) $\frac{v_A}{v_D}=\frac{20}{\sqrt{{20}^2+2\times 10\times 20}}=\frac{20}{20\sqrt{2}}=\frac{1}{\sqrt{2}}$

(IV) $\frac{t_{AC}}{t_{AB}}=\frac{2t_{AB}}{t_{AB}}=2$