Question

The projectile is thrown at the equation

Answer 1

`\begin{gathered} The\text{ given equation is,} \\ h(t)=-5t^2+25t+70 \\ \text{put h=0} \\ 0=-5t^2+25t+70 \\ t^2-5t-14=0 \\ \text{Now, solve for t} \\ t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{Here, a=1,b=-5,c=-14} \\ So, \\ t=\frac{5\pm\sqrt[]{(-5)^2-4*1*(-14)}}{2*1} \\ t=\frac{5\pm\sqrt[]{25+56}}{2} \\ t=\frac{5\pm\sqrt[]{81}}{2} \\ t=(5\pm9)/(2) \\ t=7,\text{and -2} \\ \text{The last option is true.} \end{gathered}`