A classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 110 feet per second from an initial height of 101 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 + 110t + 101 Find the two points in time when the object is 134 feet above the ground. Round your answers to the
nearest hundredth of a second (two decimal places).
The object is 134 feet off the ground at the following times: