A projectile’s launch speed is five times its speed at maxi-mum height. Find launch angle θ

The speed of a projectile at its maximum height is solely its horizontal component because the vertical component becomes zero at the maximum height.

Let's denote:

• V0 as the launch speed
• Vh as the horizontal speed
• Vv as the vertical speed
• θ as the launch angle

From the problem, we know that V0 = 5 * Vh.

We also know from physics that:

• Vh = V0 * cos(θ)
• Vv = V0 * sin(θ)

At the maximum height, Vv = 0, so the speed of the projectile is equal to Vh.

Substituting Vh = V0 * cos(θ) into the equation V0 = 5 * Vh, we get:

V0 = 5 * V0 * cos(θ)

Dividing both sides by V0 (assuming V0 ≠ 0), we get:

1 = 5 * cos(θ)

Solving for cos(θ), we get:

cos(θ) = 1/5

Finally, to find the angle θ, we take the inverse cosine of both sides:

θ = cos^-1(1/5)

So, the launch angle θ is the inverse cosine of 1/5.