a projectile is fired from the surface of the earth with a speed of 200 m/s at an angle of 30 degrees above the horizontal. If the ground is level, what is the maximum height reached by the projectile

h = V^2(sin^2(30))/2g

Vy= sin30(200)=100m/s

t=(Vf-Vi)/a, t=(0-100)/9.81, t=10.3s

Sy=Vy(t), Sy= 100(10.3), Sy= 1019m

1019M

509.7M

To find the maximum height reached by the projectile, we need to analyze the motion in the vertical direction.

First, let's break down the initial velocity into its horizontal and vertical components. The horizontal component remains constant throughout the motion, while the vertical component changes due to gravity.

Given that the initial speed is 200 m/s and the launch angle is 30 degrees above the horizontal, we can find the initial vertical component of velocity (V_y) and the initial horizontal component of velocity (V_x) using trigonometry.

V_y = V * sin(angle)
V_x = V * cos(angle)

V = 200 m/s (given speed)
angle = 30 degrees (given angle)

Plugging in the values, we get:
V_y = 200 m/s * sin(30 degrees)
V_x = 200 m/s * cos(30 degrees)

V_y = 100 m/s
V_x = 173.2 m/s (approximately)

Now, we can use the kinematic equation in the vertical direction to find the maximum height (h_max) reached by the projectile:

V_f^2 = V_i^2 + 2 * a * d

Here, V_f is the final vertical velocity, V_i is the initial vertical velocity, a is the acceleration due to gravity (-9.8 m/s^2), and d is the displacement in the vertical direction.

At the maximum height, the final vertical velocity is zero, so V_f = 0.

0 = V_i^2 + 2 * a * d

Plugging in the values, we get:
0 = (100 m/s)^2 + 2 * (-9.8 m/s^2) * d

Simplifying the equation, we have:
0 = 10000 m^2/s^2 - 19.6 m/s^2 * d

Rearranging the equation, we find:
d = (10000 m^2/s^2) / (19.6 m/s^2)
d ≈ 510.2 meters

So, the maximum height reached by the projectile is approximately 510.2 meters.