A projectile is launched at an angle of 30 to the horizontal with a speed of 30 m/s.

How does the horizontal component of its velocity 1.0 s after launch compare with its horizontal component of velocity 2.0 s after launch?

There is no horizontal force.

Therefore there is no change in the horizontal component of velocity.

To compare the horizontal components of the projectile's velocity at different times, we need to remember that the horizontal component of velocity remains constant throughout the motion.

Given that the initial angle of the projectile is 30 degrees to the horizontal and the initial speed is 30 m/s, we can determine the horizontal component of velocity using trigonometry.

The horizontal component of velocity (Vx) can be calculated using the formula:

Vx = V * cos(theta),

where V is the initial speed and theta is the launch angle.

Plugging in the values, we have:

Vx = 30 m/s * cos(30 degrees)
Vx = 30 m/s * 0.866
Vx ≈ 25.98 m/s

Thus, the horizontal component of the projectile's velocity 1.0 s after launch is approximately 25.98 m/s.

Since the horizontal component of velocity remains constant, the horizontal component of velocity 2.0 s after launch will also be 25.98 m/s.

To compare the horizontal component of the velocity of a projectile at different times after launch, we need to understand the basic principles of projectile motion.

A projectile's motion can be divided into two components: the horizontal motion and the vertical motion. The horizontal motion is constant and unaffected by gravity, while the vertical motion is influenced by gravity and follows a parabolic trajectory.

Given that a projectile is launched at an angle of 30° to the horizontal with a speed of 30 m/s, we can break down its initial velocity into horizontal and vertical components.

The horizontal component of the initial velocity can be determined using the formula:
Vx = V * cos(θ)
where Vx is the horizontal component of the velocity, V is the initial velocity, and θ is the launch angle.

Plugging in the values from the given data:
Vx = 30 m/s * cos(30°)
Vx = 30 m/s * (√3 / 2)
Vx ≈ 26 m/s

So, the horizontal component of the initial velocity is approximately 26 m/s.

Now, let's consider the horizontal component of the velocity at two different times: 1.0 s and 2.0 s after launch.

Since the horizontal component of velocity remains constant throughout the flight, the horizontal component of velocity 1.0 s after launch would be the same as the initial horizontal component of velocity, approximately 26 m/s.

Similarly, the horizontal component of velocity 2.0 s after launch would also remain constant and would also be approximately 26 m/s.

Therefore, the horizontal component of velocity 1.0 s and 2.0 s after launch would be the same, approximately 26 m/s.