The particle will accelerate 5m/s every second until it reaches a maximum of whatever your graph/diagram goes to, I'm in physical science and this is somewhat similar to what I am doing now but I'm not sure if that was what your looking for.
Answer:
∆h = 0.071 m
Explanation:
I rename angle (θ) = angle(α)
First we are going to write two important equations to solve this problem :
Vy(t) and y(t)
We start by decomposing the speed in the direction ''y''


Vy in this problem will follow this equation =

where g is the gravity acceleration

This is equation (1)
For Y(t) :

We suppose yi = 0

This is equation (2)
We need the time in which Vy = 0 m/s so we use (1)

So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)

2.236 m is the maximum height from the shell (in which Vy=0 m/s)
Let's calculate now the height for t = 0.555 s

The height asked is
∆h = 2.236 m - 2.165 m = 0.071 m
Answer:
2.72×10^-7
Explanation:
velocity = frequency × wavelength
2.05×10^8=7.55×10^14 wavelength
wavelength = 2.05×10^8/7.55×10^14
wavelength = 2.72×10^-7