Using trigonometry, determine the maximum height of a ball to the nearest tenth (1 yard= 3ft)

To determine the maximum height of the ball, we need to use the equation of motion and trigonometry.

The equation of motion for the vertical direction of motion is given by:

y = y₀ + v₀t - 1/2gt²

where:
y is the vertical displacement,
y₀ is the initial vertical position,
v₀ is the initial vertical velocity,
g is the acceleration due to gravity (approximately -32 ft/s²),
and t is the time.

At the maximum height, the vertical velocity becomes zero (v = 0). Therefore, we have:

0 = v₀ - gt

Solving for t, we find:

t = v₀ / g

To determine the maximum height, we need to find the value of y when t = t/2. We can substitute this value of t into the equation of motion to find the maximum height:

y = y₀ + v₀(t/2) - 1/2g(t/2)²

Simplifying the equation:

y = y₀ + v₀t/2 - 1/8gt²

Since y₀ is the initial vertical position and we are assuming the ball is thrown from the ground, y₀ = 0. Also, the initial vertical velocity v₀ can be calculated using the horizontal velocity and the launch angle using trigonometry. Let's call the launch angle θ.

v₀ = v × sin(θ)

To determine the maximum height, we need the horizontal velocity v. However, the horizontal velocity v has not been provided in the problem statement. We will need this information to solve the problem further.