At its maximum height <em>y</em>, the ball has zero vertical velocity. So we have
0² - (23 m/s)² = 2 * (-<em>g</em>) * <em>y</em>
where <em>g</em> = 9.80 m/s² is the acceleration due to gravity. Solve for <em>y</em> :
<em>y</em> = (23 m/s)² / (2 * (9.80 m/s²))
<em>y</em> ≈ 27 m
Answer:
acceleration of car B appears at slowing down by 16 km/h each second
Explanation:
As we know by the concept of relative acceleration
![a_{BA} = a_B - a_A](https://tex.z-dn.net/?f=a_%7BBA%7D%20%3D%20a_B%20-%20a_A)
now we know that
Car A is moving with speed 100 km/h such that its speed is increasing at rate of 8 km/h each second
so acceleration of car A is given as
![a_A = 8 km/h/s](https://tex.z-dn.net/?f=a_A%20%3D%208%20km%2Fh%2Fs)
now car B is also moving at same speed of 100 km/h but its speed is slowing down at rate of 8 km/h each second
so acceleration of car B is negative as it is decelerating
![a_B = - 8 km/h/s](https://tex.z-dn.net/?f=a_B%20%3D%20-%208%20km%2Fh%2Fs)
So here we have from above formula
![a_{BA} = a_B - a_A = - 8 - 8 = -16 km/h/s](https://tex.z-dn.net/?f=a_%7BBA%7D%20%3D%20a_B%20-%20a_A%20%3D%20-%208%20-%208%20%3D%20-16%20km%2Fh%2Fs)
so acceleration of car B appears at slowing down by 16 km/h each second