A ball rolls off the edge of a horizontal table top 4m 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.

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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 $v = \frac{B C}{t_{A C}} = \frac{5.0}{0.9} = 5.55 m / s$

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A ball rolls off the edge of a horizontal table top 4m 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. ___

Using h=12gt2, we get hAB=12gt2AC or tAC=√2hABg √2×49.8=0.9s Further, BC=vtAC or v=BCtAC=5.00.9=5.55m/s

A ball rolls off the edge of a horizontal table top 4m 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.

___

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 $v = \frac{B C}{t_{A C}} = \frac{5.0}{0.9} = 5.55 m / s$