Balls A and B are released from rest from roof of a building at t=0 and t=2 s, respectively. The ball A strikes with ground and comes back with same speed. After some time, both balls A and B meet with each other at a height of 55 m from the ground. If the height of building is 60n meter, then the value of n is _______

Moderate

Solution

$If they meet at t = t_{0} . t_{0} i s t h e t o t a l t i m e t a k e n b y b a l l A f r o m P t o Q a n d Q t o R (t_{0} - 2) i s t h e t o t a l t i m e t a k e n b y b a l l B f r o m P t o R$
$B a l l B : P t o R ∴ h - 55 = \frac{1}{2} g {(t_{0} - 2)}^{2} (for ball B) ......(i)$
where, h is height of the building. Let say the velocity of each ball at the meeting point is v₀.
$∴ v_{0} = g (t_{0} - 2) - - - - (i i) for ball B$
Let the time taken by ball A to go from meeting point to ground is t₁.
$\begin{array}{l} t o t a l t i m e t a k e n b y A = t i m e t a k e n f r o m P t o R + t i m e t a k e n f r o m R t o Q + t i m e t a k e n f r o m Q t o R \\ t_{0} = (t_{0} - 2) + t_{1} + t_{1} \\ 2 = 2 t_{1} \\ t_{1} = 1 s - - - (i i i) \end{array}$
$R t o Q : ∴ 55 = v_{0} t_{1} + \frac{1}{2} g t_{1}^{2}$
$\begin{array}{l} o r 55 = g (t_{0} - 2) t_{1} + \frac{1}{2} g t_{1}^{2} \\ o r 5.5 = (t_{0} - 2) t_{1} + \frac{t_{1}^{2}}{2} = t_{0} - 2 + \frac{1}{2} \\ o r 5 = t_{0} - 2 \end{array}$
$\begin{array}{l} ∴ t_{0} = 7 s \\ S u b s t i t u t e i n (i) \\ ∵ h - 55 = \frac{1}{2} g {(t_{0} - 2)}^{2} \end{array}$
$\begin{array}{l} Or h - 55 = 5 (7 - 2)^{2} = 125 \\ ∴ h = 55 + 125 = 180 m = 60 n \\ ∴ n = 3 \end{array}$

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