The speed of the ball bearing is given by the first derivative of the distance function.
Using the power rule in our function, we have
[tex]\begin{gathered} s^{\prime}(t)=\frac{d}{dt}(4.9(\sin \theta)t^2)=4.9(\sin \theta)\frac{d}{dt}(t^2)=4.9(\sin \theta)(2t) \\ s^{\prime}(t)=9.8(\sin \theta)t \end{gathered}[/tex]
To complete the table we just have to evaluate the sine function on the given θ values.
[tex]\begin{gathered} \theta=0\Rightarrow s^{\prime}(t)=9.8(\sin 0)t=9.8\cdot0\cdot t=0 \\ \theta=\frac{\pi}{4}\Rightarrow s^{\prime}(t)=9.8(\sin (\frac{\pi}{4}))t=9.8\cdot\frac{\sqrt[]{2}}{2}\cdot t=4.9\sqrt[]{2}t \\ \theta=\frac{\pi}{3}\Rightarrow s^{\prime}(t)=9.8(\sin (\frac{\pi}{3}))t=9.8\cdot\frac{\sqrt[]{3}}{2}\cdot t=4.9\sqrt[]{3}t \\ \theta=\frac{\pi}{2}\Rightarrow s^{\prime}(t)=9.8(\sin (\frac{\pi}{2}))t=9.8\cdot1\cdot t=9.8t \\ \theta=\frac{2\pi}{3}\Rightarrow s^{\prime}(t)=9.8(\sin (\frac{2\pi}{3}))t=9.8\cdot\frac{\sqrt[]{3}}{2}\cdot t=4.9\sqrt[]{3}t \\ \theta=\frac{3\pi}{4}\Rightarrow s^{\prime}(t)=9.8(\sin (\frac{3\pi}{4}))t=9.8\cdot\frac{\sqrt[]{2}}{2}\cdot t=4.9\sqrt[]{2}t \\ \theta=\pi\Rightarrow s^{\prime}(t)=9.8(\sin \pi)t=9.8\cdot0\cdot t=0 \end{gathered}[/tex]
According to the table, the biggest value for the speed happens at θ = π/2.
