Question

In a fireworks display, a rocket is launched from the ground with a speed of 18.0 m/s and a direction of 51.0° above the horizontal. During the flight, the rocket explodes into two pieces of equal mass. (a) What horizontal distance from the launch point will the center of mass of the two pieces be after both have landed on the ground? (b) If one piece lands a horizontal distance of 26.0 m from the launch point, where does the other piece land?

Answer 1

Answer:38.66 m

Explanation:

Given

launch angle $\theta =51^{\circ}$

launch velocity $u=18 m/s$

center of mass continue to travel its original Path so it center of mass will be at a distance of

$R=\frac{u^2\sin 2\theta }{g}$

$R=\frac{18^2\sin 102}{9.8}$

$R=32.33 m$

Center of mass will be at $x=32.33 m$

(b)if one of the piece will be at x=26  m then other will be at

$x_{com}=\frac{m_1x_1+m_2x_2}{m_1+m_2}$

$x_{com}=\frac{\frac{m}{2}\cdot 26+\frac{m}{2}\cdot x_0}{\frac{m}{2}+\frac{m}{2}}$

$32.33=\frac{26+x_0}{2}$

$x_0=38.66 m$