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Moment of Inertia of a Rod

A body A is...

Question

A body $A$ is dropped from the top of a cliff $42 m$ high and at the same time, another body $B$ is thrown vertically upwards from the ground with a velocity of $21 m / s$ . When and where will they meet?

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Solution

$\begin{matrix} x = u t + \frac{1}{2} a t^{2} x = \frac{1}{2} \times g t^{2} . . . . . . (1) u_{2} - x = 21 t - \frac{1}{2} \times g t^{2} . . . . . . (2) u_{2} - \frac{1}{2} \times g t^{2} = 21 t - \frac{1}{2} \times g t^{2} t = 2 sec x = \frac{1}{2} \times g \times 4 = 2 \times 9.81 = 18.62 (f r o m t o p) 42 - 18.62 = 21.38 (f o r b o t t o m) \end{matrix}$

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