Answer:
Maximum height reached is given as H = 5.00 m
Step-by-step explanation:
As we know that if the ball is projected at some angle with the horizontal then the range of the projectile is given by the formula
![R = (v^2 sin2\theta)/(g)](https://img.qammunity.org/2021/formulas/physics/high-school/t6bb49gj9hbvmj46869kq4kjks32j0rv4y.png)
here we have
![R = 55 m](https://img.qammunity.org/2021/formulas/physics/high-school/a32g5x0gpve0nyhv4dl8nxl3oo55inhvgc.png)
![\theta = 20^o](https://img.qammunity.org/2021/formulas/physics/high-school/922ydk01kkqw2xoxozuclotkdsa4inp03t.png)
v = 40 m/s
Now we have
![55 = (40^2 sin40)/(g)](https://img.qammunity.org/2021/formulas/physics/high-school/bi2fu2jz32lj9gwjtp3fc06t5xrjznlsr2.png)
![g = 18.7 m/s^2](https://img.qammunity.org/2021/formulas/physics/high-school/8htx3rma3euw481g3ykomkawu545r3wil7.png)
now for maximum height we have
![H = (v^2sin^2\theta)/(2g)](https://img.qammunity.org/2021/formulas/physics/high-school/vx0rgpcik1vgji9w9wbxasheu3exxcbrci.png)
![H = (40^2 sin^220)/(2(18.7))](https://img.qammunity.org/2021/formulas/physics/high-school/jfxbgbgcnt8etojc2hsehq2m2rrj5cuhow.png)
![H = 5 m](https://img.qammunity.org/2021/formulas/physics/high-school/l6imh6p3i3ln6nzeoirvbq772lc3hwqz60.png)