A ball is dropped from the height if it takes one second to cross last $55 m$ before hitting the ground. Find the height from which it was drop.

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Solution

$\begin{matrix} B C - S = v_{1} t + \frac{1}{2} g t^{2} 55 = v_{1} \times 1 + \frac{1}{2} \times 10 \times 1 v_{1} = 50 m s^{- 1} s o v_{2} = v_{1} + g t v_{2} = 50 + 10 \times 1 = 60 m s^{- 1} A C - {v_{2}}^{2} - u^{2} = 2 g S 60^{2} = 2 \times 10 S S = 180 m \end{matrix}$

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A ball is dropped from a height. If it takes 0.2s to cross the last 6 m before hitting the ground, find the height from which it was dropped.

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