A projectile is launched over level ground at a launch angle of 70o with an initial velocity

vo. At some later time while the projectile is on its way to the peak in its trajectory, its
velocity vector makes an angle of 50o with respect to the horizontal. What is the
magnitude of the projectile’s horizontal velocity at that point?

v(x) =v₀(x) =v₀cosα =const

To find the magnitude of the projectile's horizontal velocity at the given point in its trajectory, we need to consider the components of its velocity vectors.

Let's break down the given information:

1. The launch angle is 70 degrees.
2. At some later time while the projectile is on its way to the peak, its velocity vector makes an angle of 50 degrees with respect to the horizontal.

First, we need to determine the initial vertical and horizontal velocities of the projectile.

1. The launch angle is 70 degrees. This means that the initial velocity can be broken down into two components: vertical and horizontal.
- The vertical component of velocity (v_vertical) is given by vo * sin(θ), where θ is the launch angle.
- The horizontal component of velocity (v_horizontal) is given by vo * cos(θ).

Next, we need to find the vertical and horizontal velocities of the projectile at the given point in its trajectory.

2. At the given point, the velocity vector makes an angle of 50 degrees with respect to the horizontal. Let's call this point A.
- The vertical component of velocity at point A (v_A_vertical) can be calculated by using the given angle and the initial vertical velocity: v_A_vertical = v_vertical * sin(50°).
- The horizontal component of velocity at point A (v_A_horizontal) can be calculated by using the given angle and the initial horizontal velocity: v_A_horizontal = v_horizontal * cos(50°).

Finally, we can find the magnitude of the projectile's horizontal velocity at point A by taking the absolute value of the horizontal component of velocity.

Magnitude of horizontal velocity at point A = |v_A_horizontal|

Remember to substitute the values of vo, θ, and the given angles to compute the desired result.

This is not a place to dump your homework. I doubt you will get much help unless you start showing some work on your part. And Ms Sue is right. Using umpteen screen names only makes it less likely you will get help.