A jet plane comes in for a downward dive as shown in the figure below.

Result:

,

Explanation:

The electromagnetic attraction between the electron and the proton in the nucleus is equivalent to the centripetal force:

where

k represents the Coulomb constant

e denotes the charge of the electron

e denotes the charge of the proton in the nucleus

r signifies the distance from the electron to the nucleus

v indicates the velocity of the electron

is the mass of the electron

Rearranging for v, we determine

Inside a hydrogen atom, the distance separating the electron from the nucleus is roughly

while the mass of the electron is

and the charge is

By plugging in the values into the formula, we achieve

Response:

y= 240/901 cos 2t+ 8/901 sin 2t

Clarification:

To determine mass m=weight/g

  m=8/32=0.25

To calculate the spring constant

Kx=mg    (with c=6 inches and mg=8 pounds)

K(0.5)=8               (6 inches converts to 0.5 feet)

K=16 lb/ft

The governing equation for the spring-mass system is

my''+Cy'+Ky=F  

Inserting the known values yields

0.25 y"+0.25 y'+16 y=4 cos 20 t  ----(1) (given C=0.25 lb.s/ft)

Assuming the steady state equation for y is

y=A cos 2t+ B sin 2t

To determine constants A and B, we must equate this with equation 1.

Next, we find y' and y" by differentiating with respect to t.

y'= -2A sin 2t+2B cos 2t

y"=-4A cos 2t-4B sin 2t

Now, substitute the values of y", y' and y into equation 1

0.25 (-4A cos 2t-4B sin 2t)+0.25(-2A sin 2t+2B cos 2t)+16(A cos 2t+ B sin 2t)=4 cos 20 t

By comparing coefficients on both sides

30 A+ B=8

A-30 B=0

From this, we find

A=240/901 and B=8/901

Thus, the steady state response

y= 240/901 cos 2t+ 8/901 sin 2t

Let's consider a few possibilities. 1. The lowest velocity of the paratrooper would be just before hitting the ground. 2. Given that the jump originated from a relatively short height, the paratrooper utilized a static line, allowing the parachute to deploy almost instantly after leaping. Hence, we will convert 100 mi/h to ft/s: 100 mi/h * 5280 ft/mi / 3600 s/h = 146.67 ft/sec. Based on the first assumption, the maximum distance fallen by the paratrooper would equate to 8 seconds at 146.67 ft/s, translating to 8 s * 146.67 ft/s = 1173.36 ft. This calculated distance is nearly on par with the jump height, validating both assumptions 1 and 2. Thus, this scenario seems plausible. Moreover, considering the terminal velocity for a parachutist in a freefall position with limbs spread out typically reaches 120 mi/h, which is slightly above the 100 mi/h mentioned in the article. This as well aligns with the notion of the parachute acting like a flag, adding some air resistance.

When the boat submerges completely in the pond, the water level of the pond rises.