Answer:
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
Step-by-step 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