The speed of the object can be given as,
[tex]v=\frac{d}{t}[/tex]Also, the final speed of the object can be expressed as,
[tex]v=u+at[/tex]For marble, the distance covered is 10 m in 1.01 s. Plug in the known values,
[tex]\begin{gathered} v=\frac{10\text{ m}}{1.01\text{ s}} \\ =9.90\text{ m/s} \end{gathered}[/tex]The acceleration can be calculated as,
[tex]\begin{gathered} 9.90\text{ m/s=0 m/s+a(1.01 s)} \\ a=\frac{9.90\text{ m/s}}{1.01\text{ s}} \\ =9.80m/s^2 \end{gathered}[/tex]For bowling ball, the distance covered is 10 m in 1.03 s.
Substitute the known values in the formula of speed.
[tex]\begin{gathered} v=\frac{10\text{ m}}{1.03\text{ s}} \\ \approx9.71\text{ m/s} \end{gathered}[/tex]Plug in the known values in formula of final speed to calculate the acceleration.
[tex]\begin{gathered} 9.71\text{ m/s=0 m/s+a(1.03 s)} \\ a=\frac{9.71\text{ m/s}}{1.03\text{ s}} \\ \approx9.43m/s^2 \end{gathered}[/tex]For baseball the distance covered is 10 m in 0.99 s.
Substitute the known values to calculate the speed of baseball.
[tex]\begin{gathered} v=\frac{10\text{ m}}{0.99\text{ s}} \\ \approx10.1\text{ m/s} \end{gathered}[/tex]Substitute the known values in the expression of final speed to obtain the acceleration.
[tex]\begin{gathered} 10.1\text{ m/s=0 m/s+a(0.99 s)} \\ a=\frac{10.1\text{ m/s}}{0.99\text{ s}} \\ \approx10.2m/s^2 \end{gathered}[/tex]Therefore, the speed and the acceleration of the baseball is maximum which is 10.1 m/s and 10.2 m/s2 respectively, because the time taken by ball to cover the distace is minimum.
