Question

During a physics​ experiment, a class drops a golf ball off a bridge toward pavement below. The bridge is 7777 ft high. The function hequals=negative 16 t squared plus 77−16t2+77 gives the golf​ ball's height h in feet above the pavement after t seconds. Use the graph of the function shown here. After how many seconds does the golf ball hit the​ pavement?

Answer 1

Set h = 0 and solve: 0 = −16t^2 + 77 ⇒ t^2 = 77/16 ⇒ t = ±√(77)/4.
Take the positive root: t = √77 / 4 ≈ 2.19 seconds.
So the ball hits the pavement after about 2.19 s.