A model jet is fired up in the air from a 16-foot platform with an initial upward velocity of 52 feet per second. The height of the jet above ground after t seconds is given by the equation -16t^2 + 52t + 16, where h is the height of the jet in feet and t is the time in seconds since it is launched. What is the maximum height the jet reaches, to the nearest foot? A. 777 feet B. 58 feet C. 30 feet D. 16 feet

Question

Eiwoh2 · Answer

@Ccallander

sillybilly123 · Answer

Eqn missing :(

Mercury · Answer

old question but will answer so this can be closed

there are a couple of ways you could approach this
1. graphing and looking for the vertex
2. calculating the vertex manually by x = -b/(2a) and plugging this x value back into the equation
3. if you know any calculus, you can take the derivative wrt x and solve for the x-coordinate of the vertex that way, then plug back into the equation

Mercury · Answer

(actually they use t, not x, so replace x with t in my previous statements)