An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the earth with a speed of 0.99543c relative to the earth. A scientist at rest on the earth's surface measures that the particle is created at an altitude of 43.0 km .
(a) As measured by the scientist, how much time does it take the particle to travel the 45.0 km to the surface of the earth?
(b) Use the length-contraction formula to calculate the distance from where the particle is created to the surface of the earth as measured in the particle’s frame.
(c) In the particle’s frame, how much time does it take the particle to travel from where it is created to the surface of the earth? Calculate this time both by the time dilation formula and from the distance calculated in part (b). Do the two results agree?

Check attachment

The question have two distance

I decided to use the one in the question "a" in attachment and I will use the other one here

Explanation:

Given that,

Speed of particle relative to the earth is

V = 0.99543c

Where c is speed of light

c = 3 × 10^8 m/s

Particle height as detected by scientist is 43km

The initial length is 43km

Lo = 43km

Lo = 43,000m

A. Time taken for the particle to reach the earth surface?

Speed = distance / Time

Time = distance / speed

t = L / V

t = 43,000 / 0.99543c

t = 43,000 / (0.99543 × 3 × 10^8)

t = 1.4399 × 10^-4 seconds

b. Initial Lenght is given as Lo = 43km

Using length contraction formula

L = Lo√(1 — u² / c²)

L = 43√[1 — (0.99543c)² / c²]

L = 43√[1 — 0.990881c² / c²]

L = 43√[1 — 0.990881]

L = 43 × √(9.1191 × 10^-3)

L = 43 × 0.095494

L = 4.1062 km

c. Using time dilation formula

∆to = ∆t√(1 — u² / c²)

∆t is gotten from question a

∆t = 1.4399 × 10^-4 seconds

∆to = ∆t√[1 — (0.99543c)² / c²]

∆to = ∆t√[1 — 0.990881c² / c²]

∆to = ∆t√[1 — 0.990881]

∆to = ∆t × √(9.1191 × 10^-3)

∆to = 1.4399 × 10^-4 × 0.095494

∆to = 1.375 × 10^-5 seconds

To check if the time dilation agree

t = L / V

t = 4.1062 × 1000 / 0.99543c

t = 4.1062 × 1000 / 0.99543 × 3 × 10^8

t = 1.375 × 10^-5 seconds

The time dilation agreed

the maxium height is 34m/s