Please help. 48 points and the first answer is brainliest!
An object is launched upwards from an initial height of 4 feet above ground, with an initial velocity of 48 feet per second. If the object's position at time X is given by
f(x) = −16x^2 + vt + s, where V is initial velocity and S is initial height, which equation can be used to find the maximum height of the object after x seconds?
A) f(x) = −16(x − 3)^2 + 40
B) f(x) = −16(x − 1.5)^2 + 40
C) f(x) = −16(x − 1.5)^2 + 20
D) f(x) = −16(x − 1.5)^2 + 32