The launching speed of a certain projectile is 9.8 times the speed it has at its maximum height. Calculate the elevation angle at launching.

To calculate the elevation angle at launching, we need to use some trigonometric concepts and equations.

Let's assume that the launching speed of the projectile is "v" m/s. According to the given condition, the speed at maximum height is 1/9.8 times the initial speed, so it is v/9.8 m/s.

At the maximum height, the vertical component of the projectile's velocity becomes zero, as the projectile reaches its peak. The horizontal component of velocity remains constant throughout the motion.

Now, let's break down the initial velocity "v" into its horizontal and vertical components. The horizontal component remains constant, so we'll represent it as "v_x".

The vertical component of the velocity can be represented as "v_y".

At the maximum height, the vertical component becomes zero, but we can determine its value based on the given information. The initial vertical velocity can be represented as "v_y0" (at the launch).

Using these variables, we can write the following equations:

v_x = v cosθ, where θ is the angle of elevation.
v_y0 = v sinθ.

Since we know that the speed at the maximum height is v/9.8 m/s, we can set up another equation:

v/9.8 = v_y0.

Now, we have two equations with two unknowns (v_x and v_y0). We can solve these equations to find the angle of elevation, θ.

From the equation v/9.8 = v_y0, we can solve for v_y0:

v_y0 = v/9.8.

Now, substitute v_y0 into the equation v_y0 = v sinθ:

v/9.8 = v sinθ.

Divide both sides by v:

1/9.8 = sinθ.

Take the inverse sine (sin^(-1)) of both sides:

θ = sin^(-1)(1/9.8).

Using a scientific calculator, evaluate sin^(-1)(1/9.8), and you will get the angle of elevation, θ.

Therefore, the elevation angle at launching is approximately equal to the inverse sine of 1/9.8.