Answer :
3.
We can use the projectile motion equations.
Let:
[tex]\begin{gathered} h1(t)=height_{\text{ }}of_{\text{ }}the_{\text{ }}pencil \\ h2(t)=height_{\text{ }}of_{\text{ }}the_{\text{ }}ball \end{gathered}[/tex]Where:
[tex]\begin{gathered} h1(t)=v_{01}t-\frac{1}{2}gt^2 \\ h1(t)=yf-yi \\ yf-40=35t-\frac{1}{2}(9.8)t^2 \\ h2(t)=v_{02}t-\frac{1}{2}gt^2 \\ h2(t)=62.5t-\frac{1}{2}(9.8)t^2 \end{gathered}[/tex]So, we need to find when:
[tex]yf=h2(t)[/tex]so:
[tex]35t-\frac{1}{2}(9.8)t^2+40=62.5t-\frac{1}{2}(9.8)t^2[/tex]Solve for t:
[tex]\begin{gathered} 35t+40=62.5t \\ 62.5t-35t=40 \\ 27.5t=40 \\ t=\frac{40}{27.5} \\ t\approx1.45s \end{gathered}[/tex]Answer:
The ball and the pencil will have the same height after 1.45 seconds