Answer:
The ball rolled a distance of 54 m.
Explanation:
Given:
Initial velocity of the ball is, ![u=5\ m/s](https://tex.z-dn.net/?f=u%3D5%5C%20m%2Fs)
Final velocity of the ball is, ![v=7\ m/s](https://tex.z-dn.net/?f=v%3D7%5C%20m%2Fs)
Time for rolling is, ![t=9\ s](https://tex.z-dn.net/?f=t%3D9%5C%20s)
The distance of rolling is, ![S=?](https://tex.z-dn.net/?f=S%3D%3F)
First, let us find the acceleration of the ball using Newton's equation of motion as:
![v=u+at\\a=\frac{v-u}{t}\\a=\frac{7-5}{9}=\frac{2}{9}\ m/s^2](https://tex.z-dn.net/?f=v%3Du%2Bat%5C%5Ca%3D%5Cfrac%7Bv-u%7D%7Bt%7D%5C%5Ca%3D%5Cfrac%7B7-5%7D%7B9%7D%3D%5Cfrac%7B2%7D%7B9%7D%5C%20m%2Fs%5E2)
Now, displacement of the ball can be determined using the following equation of motion:
![v^2=u^2+2aS](https://tex.z-dn.net/?f=v%5E2%3Du%5E2%2B2aS)
Rewriting the above in terms of 'S', we get
![S=\frac{v^2-u^2}{2a}](https://tex.z-dn.net/?f=S%3D%5Cfrac%7Bv%5E2-u%5E2%7D%7B2a%7D)
Plug in the known values and solve for 'S'. This gives,
![S=\frac{7^2-5^2}{2\times \frac{2}{9}}\\\\S=\frac{49-25}{\frac{4}{9}}\\\\S=\frac{9\times 24}{4}\\\\S=9\times 6=54\ m](https://tex.z-dn.net/?f=S%3D%5Cfrac%7B7%5E2-5%5E2%7D%7B2%5Ctimes%20%5Cfrac%7B2%7D%7B9%7D%7D%5C%5C%5C%5CS%3D%5Cfrac%7B49-25%7D%7B%5Cfrac%7B4%7D%7B9%7D%7D%5C%5C%5C%5CS%3D%5Cfrac%7B9%5Ctimes%2024%7D%7B4%7D%5C%5C%5C%5CS%3D9%5Ctimes%206%3D54%5C%20m)
Therefore, the ball rolled a distance of 54 m.