Given
The equation of the height is
[tex]h(t)=-16t^2+20t+950[/tex]
To find
The velocity when the stone reach the ground
Explanation
When the stone reaches the ground
[tex]\begin{gathered} h(t)=0 \\ \Rightarrow-16t^2+20t+950=0 \\ \Rightarrow16t^2-20t-950=0 \\ \Rightarrow t=\frac{20\pm\sqrt{20^2-(4\times16\times(-950)}}{2\times16} \\ \Rightarrow t=\frac{20\pm247.38}{2\times16}=8.35\text{ s} \end{gathered}[/tex]
Thus the time taken to reach the ground is 8.35s . (Here only the positive value is considered)
We know the velocity is the change in distance per unit time,
Thus,
[tex]\begin{gathered} v(t)=h^{\prime}(t) \\ \Rightarrow v(t)=-32t+20 \end{gathered}[/tex]
At t=8.35 s
[tex]\begin{gathered} v(8.35)=-32\times8.35+20 \\ \Rightarrow v(8.35)=-247.2\text{ feet/s} \end{gathered}[/tex]
Conclusion
The velocity is -247.20 feet/s
