a projectile is fired at 30 degree above the horizontal with an initial speed of 600 m/s. find the direction and magnitude of the velocity after 2 seconds?

Vy = 300-9.8t

Vx = 519.6

So, at t=2,

Vy = 280.4
Vx = 519.6

|v| = √(280.4^2+519.6^2) = 590.4
θ = arctan(280.4/519.6) = 28.4°

post it.

To find the direction and magnitude of the velocity after 2 seconds for a projectile fired at 30 degrees above the horizontal with an initial speed of 600 m/s, we can break the velocity into its horizontal and vertical components.

Let's start by finding the vertical component of the velocity. We know that the initial velocity is 600 m/s and it is fired at an angle of 30 degrees above the horizontal. The vertical component of the velocity can be found using the equation:

V_vertical = V_initial * sin(angle)

V_vertical = 600 m/s * sin(30 degrees)
V_vertical = 300 m/s

Next, let's find the horizontal component of the velocity. The horizontal component of the velocity can be found using the equation:

V_horizontal = V_initial * cos(angle)

V_horizontal = 600 m/s * cos(30 degrees)
V_horizontal = 600 m/s * √3/2
V_horizontal ≈ 519.6 m/s

Now that we have both the vertical and horizontal components of the velocity, we can find the magnitude of the velocity using the Pythagorean theorem:

Velocity = √(V_horizontal^2 + V_vertical^2)
Velocity = √(519.6^2 + 300^2)
Velocity ≈ √(269,841 + 90,000)
Velocity ≈ √(359,841)
Velocity ≈ 599.87 m/s

So, after 2 seconds, the projectile's velocity will have a magnitude of approximately 599.87 m/s.

To find the direction, we can use the inverse tangent function:

Direction = arctan(V_vertical / V_horizontal)
Direction = arctan(300 m/s / 519.6 m/s)
Direction ≈ arctan(0.577)
Direction ≈ 30 degrees

Therefore, after 2 seconds, the projectile's velocity will have a direction of approximately 30 degrees above the horizontal.