Question 1130409
A model rocket is launched with an initial velocity of 296 ft/s. The height h, in feet, of the rocket t seconds after the launch is given by h=-16t^2+296t. How many seconds after the launch will the rocket be 73 ft above the ground? Round to the nearest hundredth of a second. How many solutions are there?
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h= -16t^2+296t = 73
Solve for t