We have that the instantaneous velocity of the
shuttlecock when it hits the ground is
[tex]V_{int}=\sqrt{U^2+19.6H}[/tex]
From the question we are told
Assuming the acceleration is still -9.81 m/s2, what is the instantaneous velocity of the
shuttlecock when it hits the ground? Show your work below.
Generally the equation for acceleration is mathematically given as
[tex]a=v \frac{dv}{dx}\\\\\Therefore\\\\\2ah=v^2-u^2[/tex]
Where
acceleration is still -9.81 m/s2,
Hence,
[tex]V^2-U^2=2(-9.81)*-H[/tex]
Therefore
[tex]V_{int}=\sqrt{U^2+19.6H}[/tex]
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