This question involves the concepts of projectile motion, horizontal range, and height.
The maximum height reached by the ball is "35.15 m".
First, we will find out the launch velocity of the projectile motion of the ball, by using the formula of the horizontal range:
[tex]R= \frac{u^2Sin\ 2\theta}{g}\\\\[/tex]
where,
R = horizontal range = 301.5 m
u = launch speed = ?
θ = launch angle = 25°
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]301.5\ m = \frac{u^2Sin\ 50^o}{9.81\ m/s^2}\\\\u = \sqrt{\frac{(301.5\ m)(9.81\ m/s^2)}{Sin\ 50^o}}[/tex]
u = 62.14 m/s
Now, we will find out the maximum height (H) of the projectile motion, using the following formula:
[tex]H = \frac{u^2Sin^2\theta}{2g}\\\\H =\frac{(62.14\ m/s)^2Sin^2(25^o)}{2(9.81\ m/s^2)}[/tex]
H = 35.15 m
Learn more about projectile motion here:
https://brainly.com/question/11049671?referrer=searchResults
The attached picture shows the projectile motion.
