A projectile is fired with an initial speed v0 at

t = 0. The angle between the initial velocity
v0 and the horizontal plane is �.
x
y
v0 �
yma

To determine the maximum height, we need to analyze the projectile motion equation for the vertical component. The vertical motion of a projectile can be described by the equation:

y = y0 + v0y * t - 0.5 * g * t^2

In this equation:
- y is the vertical displacement from the initial position (y0) of the projectile
- v0y is the initial vertical component of velocity
- t is the time
- g is the acceleration due to gravity (approximately 9.8 m/s^2)

Since we're interested in the maximum height, we know that at this point, the vertical displacement is zero. So we can set y = 0 and solve for t.

0 = y0 + v0y * t - 0.5 * g * t^2

Rearranging the equation, we get:

0.5 * g * t^2 - v0y * t - y0 = 0

This is a quadratic equation in terms of t. We can solve it using the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac)) / 2a

Where:
- a = 0.5 * g
- b = -v0y
- c = -y0

After finding the time t, we can substitute it back into the equation for the vertical displacement to find the maximum height, yma.

Now, to obtain v0y, the initial vertical component of the velocity, we can use trigonometry. The initial velocity v0 can be split into its horizontal and vertical components as follows:

v0x = v0 * cos(θ)
v0y = v0 * sin(θ)

Where:
- v0x is the initial horizontal component of velocity
- v0y is the initial vertical component of velocity
- θ is the angle between the initial velocity vector and the horizontal plane

Given the angle θ, we can use the above equations to calculate v0y. With the values of v0y and y0 (which is usually the initial height), we can find the maximum height yma by solving the quadratic equation for time and substituting it back into the vertical motion equation.