Question

A ball is thrown downwards with a speed of $30\text{m/s}$ from the top of a building $150\text{m}$ high and simultaneously another ball is thrown vertically upwards with a speed of $45\text{m/s}$ from the foot of the building. Find the time after which both the balls will meet. $(g=10{\text{m/s}}^2)$

• A. 1 sec
• B. 2 sec
• C. 3 sec
• D. 4 sec

Answer 1

The correct option is B 2 sec

Using second equation of motion for both the balls, we can say
$S_1=30t+5t^2$ ....(i)
$S_2=45t-5t^2$ ....(ii)
Adding (i) and (ii)
$\Rightarrow 150=75t\Rightarrow t=2\text{s}$
Alter: Relative acceleration of both is zero since both have same acceleration in downward direction
${\overrightarrow{a}}_{AB}={\overrightarrow{a}}_A-{\overrightarrow{a}}_B=g-g=0$
$(S_{rel}{)}_y=(u_{rel}{)}_{y}t+\left(\frac{1}{2}\right)(a_{rel}{)}_y{t}^2$
Here we are talking about motion of $B$ w.r.t. $A$
${\overrightarrow{V}}_{BA}=30+45=75$
$S_{BA}=V_{BA}\times t$
$t=\frac{S_{BA}}{V_{BA}}=2\text{s}$