a baseball is thrown at an angle of 25 degrees relative to the ground at a speed of 23.0 m/s. if the ball was caught 42.0 m from the thrower, how hight was the tallest spot in the ball's path?

that still didn't really help me at all

To find the highest point of the ball's path, we can break down the vertical and horizontal motion components separately.

Given:
Initial velocity (v₀) = 23.0 m/s
Launch angle (θ) = 25 degrees
Horizontal distance (x) = 42.0 m

Step 1: Find the time of flight (t)
Since the ball's vertical motion is affected by gravity, we can use the formula for time of flight:

t = (2 * v₀ * sin(θ)) / g

where g is the acceleration due to gravity (9.8 m/s²).

Substituting the given values:
t = (2 * 23.0 * sin(25)) / 9.8
t ≈ 2.05 s

Step 2: Find the maximum height (h)
To find the maximum height, we need to find the vertical component of the velocity at the highest point. We can use the formula:

v_y = v₀ * sin(θ)

Substituting the values:
v_y = 23.0 * sin(25)
v_y ≈ 9.71 m/s

Now, we can find the maximum height (h) using the formula for vertical motion:

h = (v_y²) / (2 * g)

Substituting the values:
h = (9.71²) / (2 * 9.8)
h ≈ 4.75 m

Therefore, the highest point in the ball's path was approximately 4.75 meters above the ground.

To find the height of the highest point in the ball's path, we can break down the motion into horizontal and vertical components.

First, let's calculate the time it takes for the ball to reach the catcher. We can use the horizontal component of the motion for this calculation. The horizontal speed remains constant, so we can use the equation:

distance = horizontal speed × time

In this case, the distance is 42.0 m, and the horizontal speed is given as 23.0 m/s. Rearranging the equation, we get:

time = distance / horizontal speed

time = 42.0 m / 23.0 m/s = 1.826 seconds (rounded to three decimal places)

Now, let's deal with the vertical component of the motion. We can use the equation for vertical displacement:

vertical displacement = vertical initial velocity × time + (½) × acceleration × time²

Since the ball is going up and then coming back down to the same height, the final vertical displacement will be zero. The initial vertical velocity is given by the vertical component of the initial speed, which can be calculated using trigonometry:

vertical initial velocity = initial speed × sin(angle)

In this case, the initial speed is 23.0 m/s, and the angle is 25 degrees. Thus:

vertical initial velocity = 23.0 m/s × sin(25°) = 9.852 m/s (rounded to three decimal places)

Plugging in the values, our equation becomes:

0 = (9.852 m/s) × 1.826 s + (½) × acceleration × (1.826 s)²

Simplifying:

0 = 9.852 m/s × 1.826 s + (½) × acceleration × (1.826 s)²

0 = 17.966 m + 0.913 s² × acceleration

Now, we can solve for acceleration:

-17.966 m = 0.913 s² × acceleration

acceleration = -17.966 m / (0.913 s²)

acceleration ≈ -19.659 m/s² (rounded to three decimal places)

The negative sign indicates that the acceleration due to gravity acts downward.

Finally, let's find the height of the highest point (maximum vertical displacement). We can use the vertical component equation again:

vertical displacement = vertical initial velocity × time + (½) × acceleration × time²

However, at the highest point, the vertical velocity is zero. Therefore, our equation becomes:

0 = 9.852 m/s × time + (½) × (-19.659 m/s²) × time²

Simplifying:

0 = 9.852 m/s × time - 9.8295 m/s² × time²

Rearranging the equation:

9.8295 m/s² × time² - 9.852 m/s × time = 0

Solving this equation for time using quadratic formula:

time = [-(-9.852 ± √[(-9.852)² - 4 × 9.8295 × 0]) / (2 × 9.8295)]

time ≈ 0.499 s (rounded to three decimal places)

Now, we substitute the time back into the equation for vertical displacement:

vertical displacement = 9.852 m/s × 0.499 s + (½) × (-19.659 m/s²) × (0.499 s)²

vertical displacement ≈ 2.452 m (rounded to three decimal places)

Therefore, the highest point in the ball's path was approximately 2.452 meters above the ground.

see this: http://www.jiskha.com/display.cgi?id=1254340217

Notice the vertical height portion.