PYQ NEET- Motion In A Straight Line Kinematics L-10

Question: A ball is dropped from a high rise platform at $t=0$ starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed $v$. The two balls meet at $t=18 \mathrm{~s}$. What is the value of $v$ ?#

(Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

A) $75 \mathrm{~m} / \mathrm{s}$

B) $55 \mathrm{~m} / \mathrm{s}$

C) $40 \mathrm{~m} / \mathrm{s}$

D) $60 \mathrm{~m} / \mathrm{s}$

Answer: $75 \mathrm{~m} / \mathrm{s}$#

Sol:#

From the question, we can say distance moved by $1^{\text {st }}$ ball in $18 \mathrm{~s}=$ distance moved by $2^{\text {nd }}$ ball in $12 \mathrm{~s}$.

So, distance moved by $1^{\text {st }}$ ball in $18 \mathrm{~s}$ $$ =\frac{1}{2} \times 10 \times 18^2=1620 \mathrm{~m} $$ and distance moved by $2^{\text {nd }}$ ball in $12 \mathrm{~s}$ $$ \begin{aligned} & =v t+\frac{1}{2} g t^2 \ & \therefore 1620=v(12)+\frac{1}{2} \times 10(12)^2 \ & \Rightarrow v=75 \mathrm{~m} / \mathrm{s} \end{aligned} $$