ANSWER
![1841.84\text{ m}](https://img.qammunity.org/2023/formulas/physics/high-school/zvm3uepofq7zsruunbkpfnag93s13hphyl.png)
EXPLANATION
Parameteters given:
Initial velocity = 190 m/s
To find the maximum height, we apply the formula for the maximum height of a projectile:
![H=(u^2\sin ^2\theta)/(2g)](https://img.qammunity.org/2023/formulas/physics/high-school/yjpxbuyxab8ynkiq64drtmplt3nzuz1h95.png)
where u = initial velocity
θ = angle with the horizontal
g = acceleration due to gravity = 9.8 m/s²
From the question, the projectile is fired vertically upward. This means that the projectile will make a 90° angle with the horizontal.
Therefore, we have that the maximum height of the projectile is:
![\begin{gathered} H=(190^2\cdot\sin ^2(90))/(2\cdot9.8) \\ H=1841.84\text{ m} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/high-school/usiqf46un2fdmcsxopfrw2g5p39rwfqvi9.png)