The velocity at the bottom of the swing is [tex]\mathbf{v= \sqrt{2gh -2gd}}[/tex]. The time it took the child to reach the river is 0.62 seconds.
The objective of this question is to determine:
At the bottom of the swing, there is an equal change in both the Kinetic energy as well as the potential energy and the expression can be computed as:
[tex]\mathbf{\dfrac{1}{2}mv^2 = mg (h -d)}[/tex]
[tex]\mathbf{\dfrac{v^2}{2}= g (h -d)}[/tex]
[tex]\mathbf{v^2= 2gh -2gd}[/tex]
[tex]\mathbf{v= \sqrt{2gh -2gd}}[/tex]
According to the second equation of motion;
[tex]\mathbf{S = ut +\dfrac{1}{2}at^2}[/tex]
where;
∴
[tex]\mathbf{1.9 = 0(t) +\dfrac{1}{2}(9.8) t^2}[/tex]
[tex]\mathbf{1.9 = \dfrac{1}{2}(9.8) t^2}[/tex]
[tex]\mathbf{t^2 =\dfrac{1.9\times 2}{9.8} }}[/tex]
[tex]\mathbf{t = \sqrt{\dfrac{1.9\times2}{9.8}}}[/tex]
[tex]\mathbf{t =0.62 \ s}[/tex]
Therefore, we can conclude that the velocity at the bottom of the swing is [tex]\mathbf{v= \sqrt{2gh -2gd}}[/tex]. The time it took the child to reach the river is 0.62 seconds.
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