A projectile is launched with an initial speed V0. At its highest point, the projectile's speed is V0/2 . What was the launch angle above the horizontal?

(in deg)

horizontal velocity has to be vo/2, as that is the only component of velocity at the highest peak.

V0^2=Vv^2 + (Vo/2)^2
Vv^2=3Vo^2/4

Vv= sqrt3/2 Vo

launch angle=arctan Vv/Vh= (sqrt3/2 vo)/Vo/2= sqrt3

angle= sixty degrees

To find the launch angle above the horizontal, you can use the relationship between the horizontal and vertical components of the projectile's velocity.

Let's assume that the launch angle is θ.

At the highest point of the projectile's trajectory, the vertical component of its velocity becomes zero. This means that the horizontal component of the velocity remains constant throughout the projectile's motion.

Given that the speed at the highest point is V0/2, we can write the following equation:

V0/2 = V0 * sin(θ)

Dividing both sides of the equation by V0 gives us:

1/2 = sin(θ)

To find θ, we need to find the inverse sine (or arcsine) of both sides:

θ = arcsin(1/2)

Using a calculator, we can determine that the arcsine of 1/2 is equal to 30 degrees.

Therefore, the launch angle above the horizontal is 30 degrees.

To find the launch angle above the horizontal, we can use the concept of projectile motion. In projectile motion, the vertical and horizontal components of the motion are independent of each other.

Let's break down the problem into two parts:

1. Finding the time of flight:
The time of flight is the total time it takes for the projectile to reach its highest point and then fall back to the same level. At the highest point, the vertical component of velocity becomes zero.

Using the equation v = u + at (where v is the final velocity, u is the initial velocity, a is acceleration, and t is time), let's apply it to the vertical component of velocity:
0 = V0sinθ - gt
Rearranging the equation, we get:
t = V0sinθ / g

Since the same amount of time is taken to reach the highest point and to fall back down, the total time of flight can be given as:
2 * t = 2 * (V0sinθ / g) = (2V0sinθ) / g

2. Finding the ratio of final velocity to initial velocity:
At the highest point, the vertical component of velocity becomes zero, and only the horizontal component of velocity (V0cosθ) remains. We are given that the speed at the highest point is V0/2. Let's write the equation for the speed:
(V0/2)^2 = (V0cosθ)^2

Simplifying the equation, we get:
(V0^2 / 4) = V0^2 * cos^2θ
1/4 = cos^2θ
cosθ = ±√(1/4)
cosθ = ±1/2

Since the projectile is launched above the horizontal, we take the positive value:
cosθ = 1/2

Now, we can find the launch angle by taking the arccosine of both sides:
θ = arccos(1/2)
θ ≈ 60° or 120°

Therefore, the launch angle above the horizontal is approximately 60 degrees.