We need to find the takeoff speed of LeBron James which is required to reach a height of 1.29 m.
The takeoff speed of LeBron James is 5.03 m/s.
u = Initial velocity
v = Final velocity = 0
s = Displacement = 1.29 m
a = g = Acceleration due to gravity = \(-9.81\ \text{m/s}^2\) (-ve since he is going up)
From the kinematic equations of linear motion we have
\(v^2-u^2=2as\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0-2\times(-9.81)\times 1.29}\\\Rightarrow u=5.03\ \text{m/s}\)
The takeoff speed of LeBron James is 5.03 m/s.
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