91影视

Given :

Speed of the ball : $3.0m/s$

Flat board is tilted up at an angle : $20掳$

Target is at distance :$2.50m$

Create a coordinate system by placing $+y$vertically up the inclined plane and $+x$horizontally. Now apply the kinematic equations of projectile motion, taking advantage of the fact that the $x$position coordinate at the end of motion is $+2.5m$. Gravitational acceleration is a component acting perpendicular to the plane.

$$\begin{gathered} x_1=v_0\cos 胃鈭�t=2.5m \\ 鈬�2.5=3\cos 胃鈭�t \\ 鈬�鈭�t=\frac{2.5}{3\cos 胃} \end{gathered}$$

role="math" localid="1650727699502" $$\begin{gathered} y_1=v_0\sin 胃鈭�t-\underset{paralleltoslope}{\underset{铃�}{\frac{g路\sin 20掳}{2}}}{(鈭�t)}^2=0 \\ 鈬�0=3\sin 胃鈭�t-\frac{g路\sin 20掳}{2}{(鈭�t)}^2 \end{gathered}$$

Using (1) as a substitute in the following equation, :

$$\begin{gathered} 0=\stackrel{\tan 胃=\frac{\sin 胃}{\cos 胃}}{\stackrel{铃�}{2.5\tan 胃}}-\frac{2.5^2路g路\sin 20掳}{2{\left(3\right)}^2{\cos }^2胃} \\ =22.5\tan 胃{\cos }^2胃-\frac{6.25路g路(0.342)}{2} \\ =22.5\sin 胃\cos 胃-1.069g \\ 鈬�\sin 胃\cos 胃=0.475g \\ 鈬�胃=\boxed{\mathbf{34}\mathbf{.}\mathbf{3}\mathbf{掳}} \end{gathered}$$