Respuesta :
[tex]\begin{gathered} h=0.5m \\ g=17m/s^2 \\ m=0.1\operatorname{kg} \\ Question\text{ 1} \\ v^2_f=2gh \\ v_f=\sqrt{2gh} \\ v_f=\sqrt{(2)(17m/s^2)(0.5m)} \\ v_f=4.12\text{ m/s} \\ \text{The sp}eed\text{ is 4.12m/s} \\ \\ \text{Question 2} \\ v_f=gt \\ \text{Solving t} \\ t=\frac{v_f}{g} \\ t=\frac{4.12\text{ m/s}}{17m/s^2} \\ t=0.24s \\ \text{The time in air is 0.24s} \\ \\ \text{Question 3} \\ P=mgh \\ P=(0.1kg)(17m/s^2)(0.5m) \\ P=0.85J \\ \text{The initial potential energy is 0.85Joules} \\ \\ \text{Question 4} \\ K=\frac{mv^2_f}{2} \\ K=\frac{(0.1\operatorname{kg})(\text{ 4.12m/s})^2}{2} \\ K=0.85\text{ J} \\ \text{The kinetic energy is 0.85Joules} \end{gathered}[/tex]



