A projectile is launched from the earth’s surface at initial speed v0 at angle θ0 with the horizontal. When the projectile is at its maximum height h, it has half the speed it had when it was at half its maximum height (h/2). At what angle was the projectile launched?

67.8 degrees

To find the angle at which the projectile was launched (θ0), we need to use the given information about the projectile's speed at different heights. Let's break down the problem step by step:

Step 1: Determine the initial vertical and horizontal components of the projectile's velocity.

The initial velocity of the projectile can be decomposed into two components: the vertical component (v₀y) and the horizontal component (v₀x). Using trigonometry, we can determine these components as follows:

v₀y = v₀ * sin(θ₀)
v₀x = v₀ * cos(θ₀)

Step 2: Find the projectile's velocity at maximum height (v_max).

At maximum height, the velocity of the projectile in the vertical direction (v_y) becomes zero (v_y = 0). We can use this information to find the maximum height (h) and the time it takes for the projectile to reach that height (t_max).

Using the equation of motion for vertical motion:
v_y = v₀y - g * t_max = 0

Solving for t_max:
t_max = v₀y / g

Using the equation of motion for vertical motion:
h = v₀y * t_max - (1/2) * g * t_max²

Substituting the value of t_max:
h = v₀y² / (2g)

Step 3: Find the velocity of the projectile at half its maximum height (v_(h/2)).

At half of the maximum height (h/2), the projectile's speed is half of what it was initially. Using this information, we can find the velocity (v_(h/2)):

v_(h/2) = (1/2) * v₀

Step 4: Apply the concept of conservation of energy.

Using the conservation of energy principle, we can equate the kinetic energy (KE) at maximum height and at half of the maximum height:

(1/2) * m * v_max² = (1/2) * m * v_(h/2)²

Since mass (m) cancels out, we can simplify the equation:

v_max² = v_(h/2)²

Step 5: Substitute the values and solve for the angle (θ₀).

Substituting the expressions for v_max and v_(h/2) from step 2 and step 3, respectively, we get:

(v₀y)² / (2g) = ((1/2) * v₀ * sin(θ₀))²

Simplifying the equation:

4 * (v₀y)² = v₀² * sin²(θ₀)

Dividing both sides by v₀²:

4 * (v₀y)² / v₀² = sin²(θ₀)

Using the trigonometric identity sin²(θ) + cos²(θ) = 1, we can rewrite the equation as:

4 * (v₀y)² / v₀² = 1 - cos²(θ₀)

Rearranging the equation:

cos²(θ₀) = 1 - 4 * (v₀y)² / v₀²

Taking the square root of both sides:

cos(θ₀) = √(1 - 4 * (v₀y)² / v₀²)

Finally, solving for θ₀:

θ₀ = arccos(√(1 - 4 * (v₀y)² / v₀²))

Using the given information about the projectile's speed at different heights, substitute the values for v₀, v_(h/2), and h to calculate θ₀.