Two balls $A$ and $B$ are placed at the top of $180 m$ tall tower. Ball $A$ is released from the top at $t = 0 s$ . Ball $B$ is thrown vertically down with an initial velocity $u$ at $t = 2 s$ . After a certain time, both balls meet $100 m$ above the ground. Find the value of $u$ in $m s^{- 1}$ [ use $g = 10 m / s^{2}$ ]

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Solution

Let us assume that they meet at $t = t_{0}$

$A : 80 = \frac{1}{2} g t_{0}^{2} . . . . (i)$

$B : 80 = u (t_{0} - 2) + \frac{1}{2} g (t_{0} - 2)^{2} . . . . (i i)$

From (i), $t_{0} = 4$

$\Rightarrow 80 = 2 u + 5 (2)^{2}$

$\Rightarrow u = 30 m / s$

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Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at t=0 s. Ball B is thrown vertically down with an initial velocity u at t=2 s. After a certain time, both balls meet 100 m above the ground. Find the value of u in ms−1 [ use g=10 m/s2 ]

Let us assume that they meet at t=t0 A:80=12gt20 ....(i) B:80=u(t0−2)+12g(t0−2)2 ....(ii) From (i), t0=4 ⇒80=2u+5(2)2 ⇒u=30 m/s

Let us assume that they meet at $t = t_{0}$

$A : 80 = \frac{1}{2} g t_{0}^{2} . . . . (i)$

$B : 80 = u (t_{0} - 2) + \frac{1}{2} g (t_{0} - 2)^{2} . . . . (i i)$

From (i), $t_{0} = 4$

$\Rightarrow 80 = 2 u + 5 (2)^{2}$

$\Rightarrow u = 30 m / s$