A ball is thrown in the air from a platform that is 48 feet above the ground with an initial vertical velocity of 32 feet per second. The height of the ball, in feet, can be represented by the function h(t) = -16t^2 +32t +48, where t is the time, in seconds, since the balls was thrown. What is the maximum height of the ball?
+36916 h(t) = -16t^2 +32t +48 Max will occur at t = - b/2a = - 32/(2 *(-16)) = 1
use this value of 't' in the equation to calculate the max height attained.....
