1. **Problem statement:** A stone is thrown vertically upward with an initial velocity of $40\,ms^{-1}$ near the surface of a planet where gravitational acceleration is $8\,ms^{-2}$. (a) Find the time taken to return to the starting point. (b) Find the velocity when it returns to the starting point. 2. **Formulas and rules:** - Use the equations of motion under constant acceleration: $$v = u - gt$$ $$s = ut - \frac{1}{2}gt^2$$ where $u$ is initial velocity, $v$ is final velocity, $g$ is acceleration due to gravity, $t$ is time, and $s$ is displacement. - Upward direction is positive, gravity acts downward. 3. **Part (a) Time to return:** At the starting point, displacement $s=0$. Using: $$0 = ut - \frac{1}{2}gt^2$$ Substitute $u=40$, $g=8$: $$0 = 40t - 4t^2$$ Factor: $$t(40 - 4t) = 0$$ Solutions: $$t=0$$ (start time) or $$40 - 4t=0 \Rightarrow t=10\,s$$ So, time to return is $10$ seconds. 4. **Part (b) Velocity on return:** Velocity at time $t$: $$v = u - gt = 40 - 8 \times 10 = 40 - 80 = -40\,ms^{-1}$$ Negative sign means velocity is downward. **Final answers:** (a) Time to return: $10$ seconds. (b) Velocity on return: $-40\,ms^{-1}$ (downward).