1. The problem involves understanding the kinematics of a tennis ball tossed straight up. 2. The acceleration of the ball during its flight is constant and equal to the acceleration due to gravity, which is approximately $9.8\ \text{m/s}^2$ downward. 3. The velocity changes over time, starting from the initial speed upwards, decreasing to zero at the maximum height, then increasing downward. 4. The initial speed is the speed at which the ball is thrown upwards. 5. The maximum height is reached when the velocity becomes zero. 6. The key formula for acceleration is: $$a = \frac{\Delta v}{\Delta t}$$ where $a$ is acceleration, $\Delta v$ is change in velocity, and $\Delta t$ is change in time. 7. For velocity at any time $t$: $$v = v_0 - g t$$ where $v_0$ is initial velocity, $g$ is acceleration due to gravity ($9.8\ \text{m/s}^2$), and $t$ is time. 8. For maximum height $h$: $$h = v_0 t - \frac{1}{2} g t^2$$ where $t$ is the time to reach maximum height. 9. At maximum height, velocity $v=0$, so: $$0 = v_0 - g t \implies t = \frac{v_0}{g}$$ 10. Substitute $t$ into height formula: $$h = v_0 \cdot \frac{v_0}{g} - \frac{1}{2} g \left(\frac{v_0}{g}\right)^2 = \frac{v_0^2}{g} - \frac{1}{2} \frac{v_0^2}{g} = \frac{v_0^2}{2g}$$ Final answers: - Acceleration $a = 9.8\ \text{m/s}^2$ downward - Velocity changes as $v = v_0 - g t$ - Maximum height $h = \frac{v_0^2}{2g}$