Rock Velocity Time Ae4325

1. **State the problem:** A rock is thrown vertically upward with an initial velocity of $20$ m/s. It eventually falls into a hole $10$ m deep. We need to find: a. The velocity of the rock as it hits the bottom of the hole. b. The total time the rock is in the air until it hits the bottom. 2. **Known values and formula:** - Initial velocity, $v_0 = 20$ m/s (upward) - Displacement from the release point to the bottom of the hole, $s = -10$ m (downward, so negative) - Acceleration due to gravity, $g = 9.8$ m/s$^2$ (downward) We use the kinematic equations for uniformly accelerated motion: $$v^2 = v_0^2 + 2as$$ $$v = v_0 + at$$ where $v$ is the final velocity, $a$ is acceleration, $s$ is displacement, and $t$ is time. 3. **Part (a): Find the velocity when the rock hits the bottom of the hole.** - Here, $a = -g = -9.8$ m/s$^2$ (taking upward as positive, so gravity is negative) - Total displacement $s = -10$ m (downward) Apply the formula: $$v^2 = v_0^2 + 2as = (20)^2 + 2(-9.8)(-10) = 400 + 196 = 596$$ Calculate $v$: $$v = -\sqrt{596} \approx -24.4 \text{ m/s}$$ (The negative sign indicates velocity is downward as it hits the bottom.) 4. **Part (b): Find the total time in the air until it hits the bottom.** Use: $$v = v_0 + at$$ Substitute values: $$-24.4 = 20 - 9.8t$$ Rearranged: $$-24.4 - 20 = -9.8t$$ $$-44.4 = -9.8t$$ Divide both sides by $-9.8$: $$t = \frac{\cancel{-44.4}}{\cancel{-9.8}} = 4.53 \text{ seconds}$$ 5. **Final answers:** - Velocity at bottom: approximately $-24.4$ m/s (downward) - Total time in air: approximately $4.53$ seconds This means the rock hits the bottom of the hole moving downward at about $24.4$ m/s after about $4.53$ seconds from release.