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

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Solution

The correct option is B 2 sec
Using second equation of motion for both the balls, we can say
$S_{1} = 30 t + 5 t^{2}$ ....(i)
$S_{2} = 45 t - 5 t^{2}$ ....(ii)
Adding (i) and (ii)
$\Rightarrow 150 = 75 t \Rightarrow t = 2 s$
Alter: Relative acceleration of both is zero since both have same acceleration in downward direction
${\to a}_{A B} = {\to a}_{A} - {\to a}_{B} = g - g = 0$
$(S_{r e l})_{y} = (u_{r e l})_{y} t + (\frac{1}{2}) (a_{r e l})_{y} t^{2}$
Here we are talking about motion of $B$ w.r.t. $A$
${\to V}_{B A} = 30 + 45 = 75$
$S_{B A} = V_{B A} \times t$
$t = \frac{S_{B A}}{V_{B A}} = 2 s$

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A ball is thrown downwards with a speed of 30 m/s from the top of a building 150 m high and simultaneously another ball is thrown vertically upwards with a speed of 45 m/s from the foot of the building. Find the time after which both the balls will meet. (g=10 m/s2)

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