The speed of the ball and the distance the ball covers in the given time is required.
The speed of the ball is 29.43 m/s
The distance the ball covered is 44.15 m
Kinematics in one dimension
u = Initial velocity = 0
v = Final velocity
t = Time = 3 s
a = g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
From the Kinematic equations we have
[tex]v=u+at\\\Rightarrow v=0+9.81\times 3\\\Rightarrow v=29.43\ \text{m/s}[/tex]
[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{29.43^2-0}{2\times 9.81}\\\Rightarrow s=44.15\ \text{m}[/tex]
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