1. **Problem Statement:** A ball is thrown vertically upward with an initial speed of 25.0 m/s. We need to find: (a) How high does it rise? (b) How long does it take to reach its highest point? (c) How long does the ball take to hit the ground after it reaches its highest point? (d) What is its velocity when it returns to the level from which it started? 2. **Relevant Formulas and Concepts:** - Use the kinematic equations for uniformly accelerated motion with acceleration due to gravity $g = 9.80\ \mathrm{m/s^2}$ downward. - At the highest point, velocity $v = 0$. - Equations: - $v = v_0 - g t$ - $x = v_0 t - \frac{1}{2} g t^2$ - Time to reach max height: $t_{up} = \frac{v_0}{g}$ - Maximum height: $h = \frac{v_0^2}{2g}$ 3. **Step-by-step Solution:** **(a) Maximum height:** $$ h = \frac{v_0^2}{2g} = \frac{(25.0)^2}{2 \times 9.80} = \frac{625}{19.6} = 31.8878\ \mathrm{m} $$ **(b) Time to reach highest point:** $$ t_{up} = \frac{v_0}{g} = \frac{25.0}{9.80} = 2.551\ \mathrm{s} $$ **(c) Time to fall back down from highest point:** - The time to fall from rest at height $h$ is the same as time to rise, so: $$ t_{down} = t_{up} = 2.551\ \mathrm{s} $$ **(d) Velocity when returning to starting level:** - By symmetry, the velocity magnitude equals initial speed but directed downward: $$ v = -v_0 = -25.0\ \mathrm{m/s} $$ 4. **Summary:** - Maximum height: approximately 31.9 m - Time to reach max height: approximately 2.55 s - Time to fall back down: approximately 2.55 s - Velocity on return: -25.0 m/s (downward) This completes the solution for the first problem found in your message.