5. a baseball player throws a ball straight up and then catches it 2.4 s later at the same height from which…

Explanation:

Step1: Determine the time to reach maximum - height

The total time of flight $T = 2.4\ s$. The time to reach the maximum - height $t$ is half of the total time of flight. So $t=\frac{T}{2}=\frac{2.4}{2}=1.2\ s$.

Step2: Calculate the initial velocity using the kinematic equation

We use the kinematic equation $v = v_0−gt$, where at the maximum - height $v = 0$. Rearranging for $v_0$ (initial velocity), we get $v_0=gt$. Given $g = 9.8\ m/s^2$ and $t = 1.2\ s$, then $v_0=9.8\times1.2 = 11.76\ m/s$.

Step3: Calculate the maximum height using the kinematic equation

We use the kinematic equation $v^{2}-v_{0}^{2}=-2gh$. At the maximum - height $v = 0$. Rearranging for $h$ (height), we get $h=\frac{v_{0}^{2}}{2g}$. Substituting $v_0 = 11.76\ m/s$ and $g = 9.8\ m/s^2$, we have $h=\frac{(11.76)^{2}}{2\times9.8}=\frac{138.2976}{19.6}=7.056\ m$.

Answer:

(a) $11.76\ m/s$ (b) $7.056\ m$