A rocket is launched from a tower. The height of the rocket ,y in feet, is related to the time after launch, x in seconds , by the given equation Using this equation, find the time that the rocket will hit the ground , to the nearest 100th of second. y=-16x^2+172x+146

To find the time that the rocket will hit the ground, we need to find the value of x when y = 0. This is because the height of the rocket is given by the equation y = -16x^2 + 172x + 146, and when the rocket hits the ground, its height is 0.<br /><br />So, we set y = 0 and solve for x:<br /><br />0 = -16x^2 + 172x + 146<br /><br />This is a quadratic equation in the form of ax^2 + bx + c = 0, where a = -16, b = 172, and c = 146. We can solve this equation using the quadratic formula:<br /><br />x = (-b ± √(b^2 - 4ac)) / (2a)<br /><br />Plugging in the values of a, b, and c, we get:<br /><br />x = (-172 ± √(172^2 - 4(-16)(146))) / (2(-16))<br /><br />Simplifying further, we have:<br /><br />x = (-172 ± √(29584 + 9344)) / (-32)<br /><br />x = (-172 ± √39928) / (-32)<br /><br />x = (-172 ± 199.72) / (-32)<br /><br />Now, we have two possible solutions for x:<br /><br />x1 = (-172 + 199.72) / (-32) ≈ -1.34<br />x2 = (-172 - 199.72) / (-32) ≈ 10.62<br /><br />Since time cannot be negative, we discard the negative solution and take the positive one. Therefore, the rocket will hit the ground after approximately 10.62 seconds.