a ball is thrown into the air with an upward velocity of 48ft/s. Its height h in feet after t seconds is given by the function h=-16t^2+48t+6. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. What is the balls maximum height?
The height of the ball after t seconds is given by the equation h = -16t^2 + 48t + 6.
To find the maximum height, we need to find the vertex of the quadratic equation. The vertex of a quadratic equation of the form ax^2 + bx + c is given by the formula x = -b / (2a), where a is the coefficient of the x^2 term, b is the coefficient of the x term, and c is the constant term.
In this case, a = -16, b = 48, and c = 6. Therefore, the time at which the ball reaches its maximum height is given by t = -48 / (2*-16) = -48 / -32 = 1.5 seconds. Since time cannot be negative in this context, we can conclude that the ball reaches its maximum height after 1.5 seconds.
To find the maximum height, substitute t = 1.5 into the equation h = -16t^2 + 48t + 6. The maximum height is given by h = -16(1.5)^2 + 48(1.5) + 6 = -16(2.25) + 72 + 6 = -36 + 72 + 6 = 42 feet.
Therefore, the ball reaches its maximum height after 1.5 seconds, and the maximum height is 42 feet.