## Related questions with answers

A spaceship moving $0.80 \: \mathrm{c}$ direction away from Earth fires a missile that the spaceship measures to be moving at $0.80 \: \mathrm{c}$ perpendicular to the ship's direction of travel. Find the velocity components and speed of the missile as measured by Earth.

Solution

VerifiedK system is system attached to Earth and K' is attached to spaceship. Let the $v=0.8 \: c$ be relative speed beetween the system and let's say that it is in x direction. In K' only speed that is measured is in y direction, $u_{\text{y}}'=0.8 \: c$. Then velocity components measured from system K (from Earth) are:

$\begin{align*} & \text{x direction:} \\ u_{\text{x}} &= \frac{u_{\text{x}}' + v }{ 1 + \frac{v}{c^2}u_{\text{x}}'} \\ u_{\text{x}} &= \frac{0 + 0.8 \: c}{1 + 0} \\ u_{\text{x}} &= \boxed{0.8 \: c}, \\ & \text{y direction:} \\ u_{\text{y}} &= \frac{u_{\text{y}}'}{\gamma \left( 1+ \frac{v}{c^2} u_{\text{x}}' \right)} \\ u_{\text{y}} &= \frac{0.8 \: c}{ \frac{1}{\sqrt{1 - (0.8)^2}} \left( 1 + 0 \right)} \\ u_{\text{y}} &= \boxed{0.48 \: c}, \\ & \text{z direction:} \\ u_{\text{z}} &= \frac{u_{\text{z}}'}{\gamma \left( 1+ \frac{v}{c^2} u_{\text{x}}' \right)} \\ u_{\text{z}} &= \frac{0 }{ \frac{1}{\sqrt{1 - (0.8)^2}} \left( 1 + 0 \right)} \\ u_{\text{z}} &= \boxed{0}, \\ \end{align*}$

and speed of projectile is:

$\begin{align*} u &= \sqrt{u_{\text{x}}^2 + u_{\text{y}}^2} \\ u &= \sqrt{(0.8 \: c)^2 + (0.48 \:c)^2} \\ u &= \boxed{0.93 \:c \: < \: c}. \end{align*}$

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