Question

a volleyball is shot from one end A, 2 m above the ground, of a field.The ball passes over the net placed in the middle of the field.If the net is 2.44 m high and the field is 18 m length,determine the initial speed and the angle of shooting of the ball at the launching point A.

Answer 1

We have next system of the equations:

\begin{cases} v \cdot \sin \alpha \cdot t = 1.44 \\ v \cdot \cos \alpha \cdot t = 9 \\ v \cdot \sin \alpha - g \cdot t = 0 \end{cases} \Rightarrow \begin{cases} \tan \alpha = \dfrac{1.44}{9} \\ v \cdot \cos \alpha \cdot t = 9 \\ v \cdot \sin \alpha \cdot t = 1.44 \end{cases} \Rightarrow \begin{cases} \alpha = \arctan \left(\dfrac{1.44}{9}\right) = 9.1{}^{\circ} \\ v = \dfrac{1.44}{\sin \alpha \cdot t} = \dfrac{9.11}{t}; \\ \dfrac{1.44}{t} - 9.8 \cdot t = 0 \end{cases} \Rightarrow \begin{cases} \alpha = 9.1{}^{\circ} \\ v = \dfrac{9.11}{t}; \\ 1.44 - 9.8 \cdot t^{2} = 0 \end{cases} \Rightarrow \begin{cases} \alpha = 9.1{}^{\circ} \\ t = 0.38 \\ v = \dfrac{9.11}{t} = \dfrac{9.11}{0.38} = 23.8 \end{cases} \end{cases}

Answer: initial speed is $23.8\ \mathrm{m/s}$ and angel is 9.1 degree (9.1 degree north of east).