Answer:
The acceleration of the ball is [tex]a_y = - 0.3672 \ m/s^2[/tex]
Explanation:
From the question we are told that
The maximum height the ball reachs is [tex]H_{max} = 42.24 \ m[/tex]
The horizontal component of the initial velocity of the ball is [tex]v_{ix} = 5.57 \ m/s[/tex]
The vertical component of the initial velocity of the ball is [tex]v_{iy} = = 16.18 m/s[/tex]
The vertically motion of the ball can be mathematically represented as
[tex]v_{fy}^2 = v_{iy} ^2 + 2 a_{y} H_{max}[/tex]
Here the final velocity at the maximum height is zero so [tex]v_{fy} = 0 \ m/s[/tex]
Making the acceleration [tex]a_y[/tex] the subject we have
[tex]a_y = \frac{v_{iy} ^2}{2H_{max}}[/tex]
substituting values
[tex]a_y = - \frac{5.57^2}{2* 42.24}[/tex]
[tex]a_y = - 0.3672 \ m/s^2[/tex]
The negative sign shows that the direction of the acceleration is in the negative y-axis