Word problems and equations

Question

A rocket fired with an upward velocity of 690 ft per second. How many seconds pass before the rocket reaches the ground (S=0). S= -16(t)(t)+690t. I have worked this problem over and over to no avail.

Answer 1

Given your equation S=0 and S(t) = -16t²+690t, we want to find the value of t such that S(t) = 0. To do this, recognize that -16t²+690t = 0 when t(-16t+690) = 0. One solution to this is t=0, which doesn't make sense as the rocket is launched at t=0. The relevant solution is found by solving -16t + 690 = 0, as -16t=-690 implies t = (-690/-16) = 43.125 seconds. Thus, t=43.125 seconds when the rocket hits the ground. Have a great day/night!