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Sagot :
Given information:
Height covered;
[tex]x=0.749\text{ m}[/tex]The third equation of motion is given as,
[tex]v^2_f=v^2_i+2ax\ldots(1)[/tex]Here, v_f is the final velocity, v_i is the initial velocity, a is the acceleration (during upward motion a=-g; g is the acceleration due to gravity) and x is the jump height.
At maximum height the final velocity v_f will be zero i.e.,
[tex]v_f=0[/tex]Therefore, from equation (1),
[tex]\begin{gathered} 0^2=v^2_i-2gx \\ v^2_i=2gx \\ v_i=\sqrt[]{2gx} \end{gathered}[/tex]Substitute 9.81 m/s² for g and 0.749 m for x,
[tex]\begin{gathered} x=\sqrt[]{2\times(9.81\text{ m/s}^2)\times(0.749\text{ m})} \\ \approx3.83\text{ m/s} \end{gathered}[/tex]Therefore, the initial velocity is 3.83 m/s.
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