The height h in feet of a ball above the ground t seconds af

The height h, in feet, of a ball above the ground t seconds after being thrown upward with a velocity of 64 ft/s is given by h = 16t2 + 64t + 8. After how many seconds will the ball be 56 ft above the ground? (Enter your answers as a comma-separated list.)

Solution

The height h, in feet, of a ball above the ground t seconds after being thrown upward with a velocity of 64 ft/s is given by h = 16t² + 64t + 8. When h = 56, we have 56 =  16t² + 64t + 8 or, 16t² - 64 t + 56 - 8 = 0 or, 16t² - 64 t + 48 = 0, or on dividing both the sides by 16, we have t² - 4t + 3 = 0 or t² - 3t -t + 3 = 0 or, t(t -3) -1(t -3) = 0 or ( t -3)(t -1) = 0. Therefore, either t -3 = 0 in which case t = 3 or t -1 = 0 in which case t = 1. Thus the ball will be 56 feet above the ground either after 1 second ( when going up) or after 3 seconds (when coming down).