An object falls 16 t 2 16t2 feet in t t seconds. You drop a rock from a bridge that is 75 feet above the water. Will the rock hit the water in 2 seconds?If so how many seconds explain

Answer:No it is still 11 feet from the water at 2 seconds2.1651 seconds  is when it will hit the waterStep-by-step explanation:h(t) = -16 t^2 +75This is the formula  for the distanceThe -16t^2 is due to gravity and the 75 is the height above the bridgeWe want to know the height a t =2h(2) = -16 (2)^2 +75h(2) =-64 +75h(2) = 11The rock will be 11 ft above the waterWhen will the rock hit the waterh = 00 = -16 t^2 +75Subtract 75 from each side-75 = -16t^2 +75-75-75 = -16t^2 Divide by -16-75/-16 =  -16t^2 /-1675/16 = t^2Take the square root of each side.  We only take the positive square root since time must be positivesqrt(75/16) =t5/4 sqrt(3) = t2.1651 seconds = t