A projectile is fired with an initial velocity of 350 feet per second at an angle of 45° with the horizontal. To the nearest foot, find the maximum altitude of the projectile. The parametric equations for the path of the projectile are

x = (350 cos 45°)t, and
y = (350 sin 45°)t - 16t2.

To find the maximum altitude of the projectile, we need to determine the highest point on its trajectory.

We know that the maximum altitude occurs when the vertical component of the velocity is zero. In other words, when the projectile is at its highest point, it will momentarily stop moving upwards before starting to fall downwards.

In the given parametric equations, the vertical component of the velocity is represented by the equation:

y = (350 sin 45°)t - 16t^2

To find the time at which the vertical velocity becomes zero, we can set dy/dt = 0.

Taking the derivative of the above equation with respect to time (t) gives:

dy/dt = (350 sin 45°) - 32t

Setting dy/dt = 0 and solving for t:
(350 sin 45°) - 32t = 0
350 sin 45° = 32t
t = (350 sin 45°) / 32

Now we can substitute this value of t back into the equation for y to find the corresponding maximum altitude:

y = (350 sin 45°) * [(350 sin 45°) / 32] - 16 * [(350 sin 45°) / 32]^2

Evaluating this equation will give us the maximum altitude of the projectile.

To find the maximum altitude of the projectile, we need to determine the highest point the projectile reaches along its trajectory. We can do this by finding when the vertical component of velocity becomes zero.

Given the parametric equations:
x = (350 cos 45°)t
y = (350 sin 45°)t - 16t^2

First, let's focus on the y-component of the projectile's path equation:
y = (350 sin 45°)t - 16t^2

The vertex of a parabolic function can be found using the formula:
t = -b / (2a)

In this case, we have:
a = -16 (the coefficient of t^2)
b = (350 sin 45°) (the coefficient of t)

Substituting the values into the formula, we get:
t = -((350 sin 45°)) / (2 * -16)

Next, we simplify the equation:
t = -((350/√2)) / -32
t = (350 * √2) / 32
t ≈ 1.947 seconds (rounded to three decimal places)

Now, substitute this value back into the equation for y to find the maximum altitude:
y = (350 sin 45°) * (1.947) - 16 * (1.947^2)
y ≈ 338.635 feet

Therefore, the maximum altitude of the projectile is approximately 339 feet (rounded to the nearest foot).

Maximum height is achieved when

Vy = (Vo cos 45)*t -g t = 0

t = (350 sin 45)/g = 7.69 s
where g = 32.2 ft/s^2

The height at that time is

Vo sin 45 t - (g/2) t^2