Answer:
9.0 m
Explanation:
Let the initial velocity be 'u'.
Given:
Final velocity is half of initial velocity.
Maximum height reached by ball (H) = 12.0 m
Acceleration of the ball is due to gravity (g) = -9.8 m/s²(Downward)
Now, first, we will find initial velocity of the ball using equation of motion given as:
![v^2=u^2+2a(\Delta y)](https://tex.z-dn.net/?f=v%5E2%3Du%5E2%2B2a%28%5CDelta%20y%29)
For maximum height, final velocity is 0 as the ball stops at the maximum height temporarily. So, ![v=0\ m/s](https://tex.z-dn.net/?f=v%3D0%5C%20m%2Fs)
Also, ![\Delta y=H=12\ m](https://tex.z-dn.net/?f=%5CDelta%20y%3DH%3D12%5C%20m)
Now, plug in all the values and solve for 'u'.
![0^2=u^2+2(-9.8)(12)\\\\u^2=235.2\\\\u=\sqrt {235.2} =15.34\ m/s](https://tex.z-dn.net/?f=0%5E2%3Du%5E2%2B2%28-9.8%29%2812%29%5C%5C%5C%5Cu%5E2%3D235.2%5C%5C%5C%5Cu%3D%5Csqrt%20%7B235.2%7D%20%3D15.34%5C%20m%2Fs)
Now, consider the motion of the ball till the velocity reaches half of initial velocity.
So, final velocity (v) = ![\frac{u}{2}=\frac{15.34}{2}=7.67\ m/s](https://tex.z-dn.net/?f=%5Cfrac%7Bu%7D%7B2%7D%3D%5Cfrac%7B15.34%7D%7B2%7D%3D7.67%5C%20m%2Fs)
Now, again using the same equation and finding the new height now. Let the new height be 'h'.
So, equation of motion is given as:
![v^2=u^2+2ah\\7.67^2=15.34^2+2\times -9.8\times h\\58.83=235.2-19.6h\\\\19.6h=235.2-58.83\\\\19.6h=176.37\\\\h=\frac{176.37}{19.6}\approx9.0\ m](https://tex.z-dn.net/?f=v%5E2%3Du%5E2%2B2ah%5C%5C7.67%5E2%3D15.34%5E2%2B2%5Ctimes%20-9.8%5Ctimes%20h%5C%5C58.83%3D235.2-19.6h%5C%5C%5C%5C19.6h%3D235.2-58.83%5C%5C%5C%5C19.6h%3D176.37%5C%5C%5C%5Ch%3D%5Cfrac%7B176.37%7D%7B19.6%7D%5Capprox9.0%5C%20m)
Therefore, the height reached by the ball when velocity is decreased to one-half of the initial velocity is 9.0 m.