91Ó°ÊÓ

The maximum height the ball should reach is $h$
The distance that the launching tube can accelerate is $d$
Since the ball is thrown upwards, the final velocity is $v_f=0m/s$
Let the initial velocity is $v_i$
The acceleration of the ball is $a$
The acceleration due to gravity is $g$
The time taken by the ball to reach height is$t$

Write the equation of motion for the ball thrown above for the maximum height h,

$$\begin{gathered} h=\frac{1}{2}g{t}^2 \\ ⇒t^2=\frac{2h}{g}.....\left(1\right) \end{gathered}$$

Write the equation of motion to determine the initial velocity of the ball,

$$\begin{gathered} v_f=v_i+at \\ 0=v_i-gt \\ {v}_i=gt......\left(2\right) \end{gathered}$$

Write the equation of motion to determine the acceleration,

localid="1648616761150" $$\begin{gathered} {\left(v_f\right)}^2-{\left(v_i\right)}^2=2ad \\ 0-{\left(v_i\right)}^2=2ad......\left(3\right) \end{gathered}$$

Plug the value$v_i$ from equation (2)
$$\begin{gathered} {\left(gt\right)}^2=2ad \\ {g}^2{t}^2=2ad......\left(4\right) \end{gathered}$$
Substitute the value of $t^2$form equation (1)

$$\begin{gathered} g^2\left[\frac{2h}{g}\right]=2ad \\ 2gh=2ad \\ ⇒a=\frac{h}{d}g \end{gathered}$$

Therefore the acceleration of the ball to earn maximum points is $\frac{h}{d}g$