A ball is thrown into the air with an upward

Question

velocity of 40 ft/s. Its height h in feet after t seconds
is given by the function h = -16h 2 + 40t + 6.

A. In how many seconds does the ball reach its
maximum height?
B. What is the ball's maximum height?

Answer 1

If your expression mean h = - 16 t² + 40 t + 6 then:

Vertex of this quadratic function have coordinates:

t = - b / 2a , h = c - b² / 4a

In this case:

a = - 16 , b = 40 , c = 6

So:

t = - b / 2a = - 40 / [ 2 ∙ ( - 16 ) ] = - 40 / - 32 = ( - 8 ) ∙ 5 / ( - 8 ) ∙ 4 = 5 / 4 = 1.25 sec

h = hmax = c - b² / 4a = 6 - 40² / [ 4 ∙ ( - 16 ) ] = 6 - 1600 / - 64 = 6 + 25 = 31 ft

Answer 2

A. V = Vo + g*t = 0,
40 + (-32)t = 0,
t = 1.25 s.

B . h = -16*1.25^2 + 40*1.25 + 6 = 31 Ft.