Question

A ball rolls off the edge of a horizontal table top 4 m high. If it strikes the floor at a point 5 m horizontally
away from the edge of the table, what was its speed at the instant it left the table?

Moderate

Solution

Using, $h = \frac{1}{2} g t^{2},$ we get
$h_{A B} = \frac{1}{2} g t_{A C}^{2}$
$or t_{A C} = \sqrt{\frac{2 h_{A B}}{g}} = \sqrt{\frac{2 \times 4}{9.8}} = 0.9 s$
Further, $B C = v t_{A C}$
or $v = \frac{B C}{t_{A C}} = \frac{5}{0.9} = 5.55 {ms}^{- 1}$

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