A ball is thrown into the air with an upward velocity of 32 ft/s. Its height h in feet after t seconds is given by the function h = −16t2 + 32t + 0. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary.

Question

A: 1s; 10ft
B: 2s; 10ft
C: 1s; 26ft
D: 1s; 58ft

Answer 1

You will need the vertex.
Don't know which method you learned, but the simplest is this:
for y = ax^2 + bx + c,
the x of the vertex is -b/(2a)
= -32/-32 = 1

after 1 second, h = -16 + 32 = 16
So the max of 16 m is obtained after 1 second

or

compete the square:
h = -16(t^2 - 2t)
= -16(t^2 - 2t + 1 - 1)
= -16( (t-1)^2 - 1)
= -16(t-1)^2 + 16
so the vertex is (1,16), same result as above

unless you have a typo, none of the given choices are correct