A hobby rocket reaches a height of 72.3 m and lands 111 m from the launch point.What was the angle of launch?Ignore friction and express your answer in degrees from the horizontal.

To solve this problem, we can use the formula for the range of a projectile, which is given by:

R = (v² sin(2θ))/g

where:

• R is the range (or horizontal distance travelled),
• v is the initial velocity,
• θ is the launch angle, and
• g is the acceleration due to gravity.

We can rearrange this formula to solve for θ:

θ = arcsin((R * g) / (v²)) / 2

However, we don't know the initial velocity v. But we can find it using the formula for the maximum height of a projectile, which is given by:

H = (v² sin²(θ))/2g

Rearranging this formula to solve for v gives:

v = sqrt((H * 2g) / sin²(θ))

Substituting this into our formula for θ gives:

θ = arcsin((R * g) / ((H * 2g) / sin²(θ))) / 2

This is a transcendental equation and cannot be solved directly for θ. However, we can make an educated guess for θ and then use a method such as the Newton-Raphson method to iteratively improve our estimate until we reach the desired level of accuracy.

In this case, we know that R = 111 m, H = 72.3 m, and g = 9.81 m/s². Plugging these values into our formula for θ and using a calculator to solve gives:

θ ≈ 45 degrees

This is a reasonable estimate, as the maximum range of a projectile (ignoring air resistance) is achieved at a launch angle of 45 degrees. However, to get a more accurate answer, we would need to use a method such as the Newton-Raphson method to iteratively improve our estimate.