A projectile is launched at a speed of 141m/s and an angle of 45 degrees. How fast is it moving at the top of its curved trajectory?

Since the object is at the top of its trajectory its change in altitude (sin velocity) should be 0.

Then you just need to calculate for the speed in the horizontal axis(which never changes in simple projectile motion).

It becomes 141 cos(pi/4)
or 141 cos (45) if you prefer degrees.

This is of course assuming that its a simple projectile motion.

To find the speed of the projectile at the top of its curved trajectory, we can analyze the horizontal and vertical components of its velocity separately using the equations of motion.

Given:
Initial velocity (V₀) = 141 m/s
Launch angle (θ) = 45 degrees (converted to radians, θ = π/4)

Let's first calculate the horizontal and vertical components of the velocity.

Horizontal component (Vx) = V₀ * cos(θ)
Vertical component (Vy) = V₀ * sin(θ)

Vx = 141 m/s * cos(π/4)
≈ 99.5 m/s

Vy = 141 m/s * sin(π/4)
≈ 99.5 m/s

Since the projectile reaches its maximum height at the top of its curved trajectory, its vertical component of velocity becomes zero (Vy = 0) at that point.

At the top of the trajectory, the speed of the projectile is equal to the magnitude of the horizontal component of the velocity.

Speed at the top of the trajectory = |Vx|
= |99.5 m/s|
= 99.5 m/s

Therefore, the projectile is moving at a speed of 99.5 m/s at the top of its curved trajectory.

To find the speed of the projectile at the top of its curved trajectory, we can use the concepts of projectile motion and trigonometry. Here's how you can calculate it:

1. Break down the initial velocity into its horizontal and vertical components. Given that the projectile is launched at an angle of 45 degrees and a speed of 141 m/s, we can calculate the initial vertical and horizontal velocities.

Vertical Component (Vy):
Vy = V * sin(θ)
Vy = 141 m/s * sin(45°)
Vy ≈ 100 m/s

Horizontal Component (Vx):
Vx = V * cos(θ)
Vx = 141 m/s * cos(45°)
Vx ≈ 100 m/s

2. Determine the vertical velocity at the topmost point of the curved trajectory. At the top of the projectile's path, the vertical velocity becomes zero, since the projectile is momentarily at the peak.

3. Find the speed at the top using the horizontal velocity. Since the horizontal velocity remains constant throughout the trajectory, the speed at the top will be the same as the initial horizontal velocity.

Therefore, the projectile is moving at approximately 100 m/s at the top of its curved trajectory.