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
![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](https://tex.z-dn.net/?f=h%20%3D%20ut%20%2B%200.5%20at%5E2%5C%5C%5C%5Ch%20%3D%200%20%2B%200.5at%5E2%5C%5C%5C%5Ch%20%3D%200.5%20at%5E2%5C%5C%5C%5Ca%20%3D%20%5Cfrac%7Bh%7D%7B0.5t%5E2%7D%20%5C%5C%5C%5Ca%20%3D%20%5Cfrac%7B2.2%7D%7B0.5%20%5Ctimes%200.096%5E2%7D%20%5C%5C%5C%5Ca%20%3D%20477.43%20%5C%20m%2Fs%5E2)
The direction of the acceleration of the ball is downwards
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Answer:
sorry but I can understand the question
Answer:
C
Explanation:
got a one hundred on the test