A projectile is launched horizontally from a cliff and has initial speed 30 m/s. The projectile will hit the ground 10 s later. What is the horizontal component of the projectile's velocity 5.0 s after launch. Assume air resistance is negligible.

there is no force in the horizontal direction (gravity is vertical), so the horizontal velocity remains constant

To find the horizontal component of the projectile's velocity 5.0 seconds after launch, we need to understand the motion of the projectile.

In this case, the projectile is launched horizontally, which means it has no initial vertical velocity. Therefore, the only force acting on the projectile is the force of gravity, which causes it to accelerate vertically downward.

Given that the projectile will hit the ground 10 seconds later, we know that the vertical displacement (height) of the projectile is equal to zero. We can use this information to determine the time it takes for the projectile to reach its maximum height.

The vertical motion of the projectile can be described using the equation:

y = y₀ + v₀y * t + 1/2 * a * t²

Since the initial vertical velocity (v₀y) is zero, and the vertical displacement (y) is zero at the maximum height, the equation simplifies to:

0 = 0 + 1/2 * a * t²

Solving for t, we find that it takes 5 seconds for the projectile to reach its maximum height.

Now, we can determine the horizontal component of the projectile's velocity 5.0 seconds after launch.

The horizontal motion of the projectile is uniform, with a constant horizontal velocity (v₀x) throughout its motion. This horizontal velocity does not change due to the absence of horizontal forces.

Therefore, the horizontal component of the projectile's velocity 5.0 seconds after launch is equal to the initial horizontal velocity (v₀x).

Since the projectile is launched horizontally from a cliff with an initial speed of 30 m/s, the horizontal component of its velocity 5.0 seconds after launch is also 30 m/s.