Hope this helps a little
initial distance up = 2
initial velocity component up = 9 sin 60 = 7.79
v = 9 sin 60 - 9.8 t
when v = 0, we are there
9.8 t = 7.79
t = .795 seconds to top
h = 2 + 7.79(.795) - 4.9(.795^2)
Answer:
![\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7B%28R%5E2%5Comega_o%7D%7B%28R%5E2%20%2B%20r%5E2%29%7D)
Explanation:
As we know that there is no external torque on the system of two disc
then the angular momentum of the system will remains conserved
So we will have
![L_i = L_f](https://tex.z-dn.net/?f=L_i%20%3D%20L_f)
now we have
![L_i = (\frac{1}{2}MR^2)\omega_o](https://tex.z-dn.net/?f=L_i%20%3D%20%28%5Cfrac%7B1%7D%7B2%7DMR%5E2%29%5Comega_o)
also we have
![L_f = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega](https://tex.z-dn.net/?f=L_f%20%3D%20%28%5Cfrac%7B1%7D%7B2%7DMR%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7DMr%5E2%29%5Comega)
now from above equation we have
![(\frac{1}{2}MR^2)\omega_o = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B2%7DMR%5E2%29%5Comega_o%20%20%3D%20%28%5Cfrac%7B1%7D%7B2%7DMR%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7DMr%5E2%29%5Comega)
now we have
![\omega = \frac{MR^2\omega_o}{(MR^2 + Mr^2)}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7BMR%5E2%5Comega_o%7D%7B%28MR%5E2%20%2B%20Mr%5E2%29%7D)
![\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}](https://tex.z-dn.net/?f=%5Comega%20%3D%20%5Cfrac%7B%28R%5E2%5Comega_o%7D%7B%28R%5E2%20%2B%20r%5E2%29%7D)
Answer:
The group of light rays is reflected back towards the focal point thereby producing a magnifying effect.
Explanation: