A ball is thrown straight up from the top of a building 144 ft tall with an initial velocity of 48 ft per second. The height
s(t) (in feet)
of the ball from the ground, at time t (in seconds), is given by
s(t) = 144 + 48t − 16t^2.
Find the maximum height attained by the ball.
why, oh why, post the same thing twice?
i made a error on the other one
To find the maximum height attained by the ball, we need to determine the vertex of the quadratic function s(t) = 144 + 48t - 16t^2.
The vertex of a quadratic function in the form f(t) = at^2 + bt + c can be found using the formula t = -b / (2a), where a, b, and c are the coefficients of the quadratic function. In this case, a = -16, b = 48, and c = 144.
Substituting these values into the formula, we get t = -48 / (2 * (-16)) = -48 / (-32) = 1.5.
So, the ball reaches its maximum height at t = 1.5 seconds.
To find the maximum height, we substitute the value of t into the equation s(t) = 144 + 48t - 16t^2.
s(1.5) = 144 + 48 * 1.5 - 16 * (1.5)^2 = 144 + 72 - 16 * 2.25 = 144 + 72 - 36 = 180 feet.
Therefore, the maximum height attained by the ball is 180 feet.