A classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 104 feet per second from an initial height of 120 feet off the ground, then the height of the projectile,h, in feet, t seconds after it's shot is given by the equation:

h=-16t^2 + 104t + 120

Find the two points in time when the object is 160 feet above the ground. Round your answers to the nearest hundredth of a second (two decimal places).

Answer: The object is 160 feet off the ground at the following times:________

Thanks!