math
A ball is thrown into the air with an upward
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?
asked by
lee

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 ftposted by Bosnian

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.
posted by Henry
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