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A ball is rolled off the edge of a swimming pool and strikes the water 2.0 m away 0.30 seconds later. What was the initial speed of the ball? its speed when it strikes the water?
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at position A the vertical velocity of the ball is zero.. when the ball strikes the water the horizontal velocity does not change as there is no horizontal acceleration.. so \(\huge{v _{x}}\)=2/0.30=20/3 m/s |dw:1375877975338:dw| let the vertical displacement of the ball be \(h\) so \(h\)=1/2\(gt^2\) when the ball strikes the horizontal velocity of the ball be \(v _{y}\) so \(v _{y}=gt\) so the velocity of the ball when it strikes water \(v=\sqrt{v _{x}+v _{y}}\)
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