Question

A rocket ship is 100m in length when measured before leaving the launching pad. when in flight, a ground observer measures its length as 75m. what is the velocity of the ship

Answer 1

Let $\bf{l_1=100m,l_2=75m}$
use formula,
$\bf{\frac{l_1}{l_2}=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}}$
here, v is the velocity of rocket ship and c is the speed of light in vaccum.

now, 100/75 = 1/√(1 - v²/c²)
4/3 = 1/√(1 - v²/c²)
squaring both sides,
16/9 = 1/(1 - v²/c²)
16(1 - v²/c²) = 9
16 - 16v²/c² = 9
16v²/c² = 7
4v² = 7c²
taking square root both sides,
4v = √7c
v = √7c/4
we know, c = 3 × 10^8 m/s
so, v = 3 × 10^8 × √7/4
v = 1.98 × 10^8 m/s

Answer 2

Let the original velocity of rocket  be  L and  observed L'

L= 100 m

L'  = 75 m

Using relative  length formula,

$\frac{L}{L'}=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\\;\\\text{Where,c=velocity of light,\;\;v=velocity of rocket}\\\;\\\frac{100}{25}=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\\;\\\frac{4}{3}=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\\;\\\frac{3}{4}=\sqrt{1-\frac{v^{2}}{c^{2}}}\\\;\\\frac{9}{16}=1-\frac{v^{2}}{c^{2}}\\\;\\\frac{v^{2}}{c^{2}}=1-\frac{9}{16}\\\;\\\frac{v^{2}}{c^{2}}=\frac{16-9}{16}\\\;\\\frac{v^{2}}{c^{2}}=\frac{7}{16}\\\;\\v^{2}=\frac{7c^{2}}{16}\\\;\\v=\frac{\sqrt{7}c}{4}\\\;$

Putting  c = 3 × 10⁸ m/s

$v=\frac{\sqrt{7}\times3\times10^{8}}{4}\\\;\\v=1.98\times10^{8}\:m/s$