91Ó°ÊÓ

The velocity contains two components, one is a horizontal component and another one is vertical.

According to Newton’s laws of motion,

$\mathit{s}\mathbf{=}\mathit{u}\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}$

Here, $s,u,t,$ and a are displacement, initial velocity, time, and acceleration respectively.

Angle, $\mathrm{θ}=60°$

Horizontal distance, $u=18m$

Vertical distance, $v=8m$

For horizontal motion, gravity is zero.

$$\begin{gathered} s_H=u\cos \mathrm{θ}t \\ t=\frac{s_H}{u\cos \mathrm{θ}} \end{gathered}$$

For the vertical motion, gravity is negative

$$\begin{gathered} s_v=u\sin \mathrm{θ}t-\frac{1}{2}g{t}^2 \\ =u\sin \mathrm{θ}\left(\frac{s_H}{u\cos \mathrm{θ}}\right)-\frac{1}{2}g{\left(\frac{s_H}{u\cos \mathrm{θ}}\right)}^2 \\ 8m=18m×\tan 60°-\frac{1}{2}9.8m/s^2{\left(\frac{18m}{u\cos {60}^{°}}\right)}^2 \\ 8m=18m×1.73-\left(\frac{6350.4}{u^2}\right) \\ \left(\frac{6350.4}{u^2}\right)=23.14 \\ u=16.56m/s \\ {s}_v=u\sin \mathrm{θ}\left(\frac{s_H}{u\cos \mathrm{θ}}\right)-\frac{1}{2}g{\left(\frac{s_H}{u\cos \mathrm{θ}}\right)}^2 \\ 8m=18m×\tan 60°-\frac{1}{2}\left(9.8m/s^2\right){\left(\frac{18m}{u\cos 60°}\right)}^2 \\ 8m=\left(31.176m\right)-\left(\frac{6,350.4}{u^2}\right) \\ \frac{6,350.4}{u^2}=23.176 \\ {u}^2=274.0075m^2/s^2 \\ u=16.55m/s \end{gathered}$$ This is the initial velocity magnitude.

The final horizontal velocity component,

$$\begin{gathered} u\cos \mathrm{θ}=16.55m/s×\cos 60° \\ =8.27m/s \end{gathered}$$

The final vertical velocity is v, from the Newton’s laws of motion,

$$\begin{gathered} v^2=u^2+2gs \\ v=\sqrt{{\left(u\sin \mathrm{θ}\right)}^2+2gs} \\ =\sqrt{\left(14.4m/s^2\right)+2×-9.8m/s^2×8m} \\ =7.1m/s \end{gathered}$$

So the magnitude of velocity when the ball just strike,

$$\begin{gathered} v_{final}=\sqrt{{\left(8.27m/s\right)}^2+{\left(7.1m/s\right)}^2} \\ =10.90m/s \end{gathered}$$

For direction,

$$\begin{gathered} \tan \mathrm{θ}=\frac{7.1}{8.27} \\ \mathrm{θ}=40.64° \end{gathered}$$

The direction is below the horizontal at $\mathrm{θ}=40.64°$.