While swinging on the wrecking ball, a student notices that, at first, the wrecking

ball is in line with her birdbath when it is at a peak in its trajectory, and in line with her front door at the lowest point in the trajectory. However, since drag is acting against the wrecking ball’s motion, 3.00 minutes later the peak of the wrecking ball’s trajectory is in line with her garden gnome, the halfway point between her birdbath and front door. How much longer will it take for the peak of the ball’s trajectory to be halfway between her garden gnome and her front door? From the time she begins to watch the wrecking ball, how long will it take for the trajectory peak to be in line with the edge of her welcome mat, which is one tenth of the distance from her front door to the bird bath?

I was able to figure out mass of ball = to 720kg and velocity of ball and her is 1.90m/s

To answer both questions, we need to first analyze the motion of the wrecking ball and determine its equation of motion. Let's assume that the wrecking ball's motion can be modeled using simple harmonic motion.

1. Moving the peak of the wrecking ball's trajectory from the birdbath to the garden gnome:

Let's denote the time at which the wrecking ball reaches its peak as t = 0. At t = 0, the wrecking ball is in line with the birdbath, and at t = 3 minutes, it is in line with the garden gnome.

The equation of motion for the wrecking ball can be written as:
y = A * cos(ωt + φ)

where:
- y is the vertical displacement from the equilibrium position,
- A is the amplitude of the motion,
- ω is the angular frequency (2π / T), where T is the time period of the motion,
- t is the time, and
- φ is the phase constant.

In our case, the equilibrium position is the midpoint between the birdbath and front door.

Given that at t = 0, the wrecking ball is in line with the birdbath (which is also the equilibrium position), we can write:
y = 0 * cos(ω * 0 + φ) = 0

Substituting t = 3 minutes and y = garden gnome (which is halfway between the birdbath and front door), we have:
garden gnome = A * cos(ω * 3 + φ)

To find the time t when the peak is halfway between the garden gnome and front door, we can equate the displacement to half the amplitude:
garden gnome / 2 = A * cos(ω * t + φ)

Solving these two equations simultaneously will give us the time t.

2. Moving the peak of the wrecking ball's trajectory from the garden gnome to the welcome mat:

In this case, we need to find the time at which the displacement is equal to one-tenth of the distance between the front door and the birdbath.

Assuming the distance between the front door and birdbath is L, the distance from the front door to the birdbath is 0.9L (0.9 times L), and the distance from the birdbath to the welcome mat is 0.1L (0.1 times L).

We can set up the equation of motion as follows:
0.1L = A * cos(ω * t' + φ)

Solve this equation to find the time t' when the peak is aligned with the welcome mat.

Note: To fully solve these problems, we would need the values for A, ω, and φ, which are not provided in the question. These values can be determined by additional information or measurements.