The magnitude of the acceleration of the ball while coming to rest is 477.43 m/s²
The direction of the acceleration of the ball is downwards
The given parameters
initial velocity of the ball, u = 0
height above the ground, h = 2.2 m
time of motion of the ball, t = 96 ms = 0.096 s
The magnitude of the acceleration of the ball while coming to rest is calculated as;
let the downwards direction of the acceleration be positive
[tex]h = ut + 0.5 at^2\\\\h = 0 + 0.5at^2\\\\h = 0.5 at^2\\\\a = \frac{h}{0.5t^2} \\\\a = \frac{2.2}{0.5 \times 0.096^2} \\\\a = 477.43 \ m/s^2[/tex]
The direction of the acceleration of the ball is downwards
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