91Ó°ÊÓ

Given :

Speed of the ball : $3.0m/s$

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

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.

$$\begin{gathered} x_1=v_0\cos \mathrm{θ}∆t=2.5m \\ ⇒2.5=3\cos \mathrm{θ}∆t \\ ⇒∆t=\frac{2.5}{3\cos \mathrm{θ}} \end{gathered}$$

role="math" localid="1650727699502" $$\begin{gathered} y_1=v_0\sin \mathrm{θ}∆t-\underset{paralleltoslope}{\underset{ÁåŸ}{\frac{g·\sin 20°}{2}}}{(∆t)}^2=0 \\ ⇒0=3\sin \mathrm{θ}∆t-\frac{g·\sin 20°}{2}{(∆t)}^2 \end{gathered}$$

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

$$\begin{gathered} 0=\stackrel{\tan \mathrm{θ}=\frac{\sin \mathrm{θ}}{\cos \mathrm{θ}}}{\stackrel{Áåž}{2.5\tan \mathrm{θ}}}-\frac{2.5^2·g·\sin 20°}{2{\left(3\right)}^2{\cos }^2\mathrm{θ}} \\ =22.5\tan \mathrm{θ}{\cos }^2\mathrm{θ}-\frac{6.25·g·(0.342)}{2} \\ =22.5\sin \mathrm{θ}\cos \mathrm{θ}-1.069g \\ ⇒\sin \mathrm{θ}\cos \mathrm{θ}=0.475g \\ ⇒\mathrm{θ}=\boxed{\mathbf{34}\mathbf{.}\mathbf{3}\mathbf{°}} \end{gathered}$$