Question

In one contest at the county fair, seen in FIGURE CP4.81, a spring-loaded plunger launches a ball at a speed of $3.0m/s$from one corner of a smooth, flat board that is tilted up at a $20°$angle. To win, you must make the ball hit a small target at the adjacent corner, $2.50m$away. At what angle $\theta$should you tilt the ball launcher?

Short answer

The ball launcher should be tilted at an angle of$\mathbf{33}\mathbf{.}\mathbf{3}\mathbf{°}$

Step-by-step solution

:  Given Information 

Given :

Speed of the ball : $3.0m/s$

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

Target is at distance :$2.50m$

:  Calculating required angle

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.

$$\begin{gathered} x_1=v_0\cos \theta ∆t=2.5m \\ \Rightarrow 2.5=3\cos \theta ∆t \\ \Rightarrow ∆t=\frac{2.5}{3\cos \theta } \end{gathered}$$

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

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

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