Question 550610
t+80 has nothing to do with anything. In fact, t is a time and 80 is a distance, so you can't add them.


We have


*[tex \LARGE 80 = \frac{1}{2}gt^2 + v_0t] (assume x_0 = 0) where g is approximately -32 ft/s^2.


You want to find the value of t such that (t,80) is the vertex of a quadratic polynomial. This occurs when t = -v0/g (recall that the vertex of a polynomial in the form ax^2 + bx + c occurs at x = -b/2a). Replace t with this expression, solve for v0, then find t.