the height of a ball is modeled by f(x)=-16x(x-3), where x is the number of seconds after the ball is hut and f(x) is the height of the ball in meters?

Answer 1

f(x) = -16x(x-3) = -16x^2 + 48x

-16x^2 is the acceleration term, usually the force of gravity operating in a negative direction take the first derivative -32x or negative 32 feet per second per second deceleration due to gravity. By saying meters, this equation mixes up the gravity in feet with gravity measures in meters.

48x is the initial velocity 48 meters per second

The constant term is zero, which is the initial height

another question might be when does the ball hit the ground? Double time, 1.5 x 2 = 3 seconds

Common problem is find the maximum height. Take the first derivative and set it equal to zero and solve for x. Then -32x +48 = 0, 32x = 48, 4x = 6, 2x = 3, x = 1 1/2 seconds

plug that back into the original equation to get the maximum height

-16x^2+48x = -16(1.5)^2 + 48(1.5) = -16(2.25)+72 = -36+72 = 36 meters

in one and a half seconds, it reaches a maximum height of 36 meters