$(i)$ We observe weightlessness in a satellite although a satellite moves under Earth's gravity. Explain.
$(ii)$ $A$ ball is thrown up with a velocity of $15 \, m s^{-1}$. What is the maximum height attained by it? $(g = 9.8 \, m s^{-2})$

(N/A) $(i)$ Weightlessness is observed in a satellite because the satellite and all objects inside it are in a state of free fall towards the Earth. Since there is no external contact force (normal force) acting on the body to oppose gravity,the apparent weight becomes zero.
$(ii)$ Given:
Initial velocity $(u) = 15 \, m s^{-1}$
Final velocity at maximum height $(v) = 0 \, m s^{-1}$
Acceleration due to gravity $(g) = -9.8 \, m s^{-2}$
Using the third equation of motion: $v^2 - u^2 = 2gh$
$0^2 - (15)^2 = 2 \times (-9.8) \times h$
$-225 = -19.6 \times h$
$h = \frac{225}{19.6} \approx 11.48 \, m$
Thus,the maximum height attained is approximately $11.48 \, m$.