Answer:
9.42 m/s
Explanation:
a) Using Newton's law of motion formula:
[tex]\theta=\frac{(\omega+\omega_o)}{2}t\\\\where \ \theta=angular\ displacement=1\ rev =2\pi, w_o=initial\ velocity\ of\ discus\\=0\ rad/s, \omega=angular\ speed\ of\ discus\ at\ release,t=time\ = 1.2\ s.\\\\Hence:\\\\2\pi=\frac{(0+\omega)}{2}(1.2)\\\\\omega=\frac{2*2\pi}{1.2} \\\\\omega=10.47\ rad/s\\[/tex]
The speed of the discus at release (v) is:
v = ωr; where r = radius of discus
diameter = 1.8 m, r = diameter / 2= 1.6 / 2 = 0.9 m
v = ωr = 10.47 * 0.9
v = 9.42 m/s
