A projectile is fired at 65.0° above the horizontal. Its initial speed is equal to 37.5 m/s. Assume that the free-fall acceleration is constant throughout and that the effects of the air can be ignored.At what time after being fired does the projectile reach this maximum height?

Vo = 37.5m/s @ 65o.

Yo = 37.5*sin65 = 34 m/s.

Y = Yo + gt.
t = (Y-Yo)/g = (0-34)/-9.8 = 3.47 s.

To find the time at which the projectile reaches its maximum height, we need to analyze the motion of the projectile.

Since the projectile experiences no air resistance, the only force acting on it is gravity. The initial velocity can be split into horizontal and vertical components. The horizontal component remains constant, while the vertical component changes due to the acceleration due to gravity.

Given:
Initial speed (v₀) = 37.5 m/s
Launch angle (θ) = 65.0°

We can find the time it takes for the projectile to reach its maximum height using the following steps:

1. Calculate the initial vertical velocity (v₀y):
v₀y = v₀ * sinθ

Substitute the given values:
v₀y = 37.5 m/s * sin(65.0°)

2. Use the equation of vertical motion to find the time of flight (t):
y = v₀y * t + (1/2) * g * t²

Since the projectile reaches its maximum height, the final vertical displacement (y) is zero.
Simplifying the equation, we get:
0 = v₀y * t - (1/2) * g * t²

Rearrange the equation to form a quadratic equation:
(1/2) * g * t² - v₀y * t = 0

3. Solve the quadratic equation for t to find the time at which the projectile reaches its maximum height. Since the quadratic equation has two solutions, we need to consider only the positive solution, as we're looking for the time after being fired:
t = [ -b + sqrt(b² - 4ac) ] / (2a)

In this case, the quadratic equation is: (1/2) * g * t² - v₀y * t = 0
So, a = (1/2) * g, b = -v₀y, and c = 0

Substitute these values into the quadratic formula and calculate t.

By following these steps, you can find the time after being fired at which the projectile reaches its maximum height.