Answer/Explanation:

To solve this problem, we can use the equations of motion for projectile motion.

First, we need to find the time it takes for the ball to reach its maximum height. We can use the equation:

vy = voy + at

where vy is the vertical velocity at the maximum height (which is zero), voy is the initial vertical velocity (which is 25 m/s * sin(40)), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time to reach the maximum height. Solving for t, we get:

0 = 25 sin(40) - 9.8t

t = 2.55 seconds

Next, we can use the equation:

y = voyt + 1/2at^2

where y is the maximum height, voy is the initial vertical velocity (which is 25 m/s * sin(40)), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time to reach the maximum height (which is 2.55 seconds). Solving for y, we get:

y = (25 sin(40))(2.55) + 1/2(-9.8)(2.55)^2

y = 32.6 meters

So the maximum height of the ball is 32.6 meters.

To find the horizontal distance traveled, we can use the equation:

x = vox t

where x is the horizontal distance, vox is the initial horizontal velocity (which is 25 m/s * cos(40)), and t is the time to reach the maximum height (which is 2.55 seconds). Solving for x, we get:

x = (25 cos(40))(2.55)

x = 52.1 meters

So the ball travels a horizontal distance of 52.1 meters before hitting the ground.