Question

An archer shoots an arrow horizontally at a target 15m away. The arrow is aimed directly at the center of the target, but it hits 52cm lower. What was the initial speed of the arrow?

Answer 1

$\begin{gathered} x_o=\text{ 0 m} \\ x_f=15\text{ m} \\ y_o=0\text{ m} \\ y_f=52\operatorname{cm}=0.52m \\ t=\text{?} \\ v_o=? \\ g=9.81m/s^2 \\ y_f-y_o=\frac{gt^2}{2} \\ \text{Solving t} \\ 2(y_f-y_o)=gt^2 \\ t^2=\frac{2(y_f-y_o)}{g} \\ t=\sqrt{\frac{2(y_f-y_o)}{g}} \\ t=\sqrt[]{\frac{2(0.52m-0m)}{9.81m/s^2}} \\ t=\sqrt[]{\frac{1.04m}{9.81m/s^2}} \\ t=0.33s \\ v_o=\frac{x_f-x_o}{t} \\ v_o=\frac{15m-0m_{}}{0.33s} \\ v_o=45.45\text{ m/s} \\ \text{The intial spe}ed\text{ is 45.45 m/s} \end{gathered}$