Question

If a ball is thrown vertically upward from the roof of 48 foot building with a velocity of 112 ft/sec, its height after t seconds is s(t)=48+112t–16t^2. What is the maximum height the ball reaches "in ft"?What is the velocity of the ball when it hits the ground (height 0) "in ft/sec"?

Answer 1

You will need the vertex of this parabola
I don't know what method you learned , so I will use the simplest way not using Calculus
for y = ax^2 + bx + c, the
x of the vertex is -b/(2a)

so for yours,
t = -112/-32 = 3.5
sub that back in to get the maximum height.

when it hits the ground s(t) = 0
0 = 48 + 112t - 16t^2
16t^2 - 112t - 48 = 0
t^2 - 7t - 3 = 0
t = (7 ± √61)/2 = appr 7.4 seconds or a negative value of t, which we will reject.