Need help on some algebra quadratic applicationsAt a carnival, a new attraction allows contestants to jump off a springboard onto a platform to be...

The equation is corrected to d(t) = -16t^2+bt+20, as the
velocity is upwards is taken as positive.

So when t= 0,
d(0) = 20 ft which is platform height.

The height to which
a person jumps is given by d(t) = -16t^2+bt+20, where t is the time duration of the
jump.

Since the bell is overhead at 20ft, it must be at 20
ft above the platform. So d(t) = 40 ft at some time t during the
jump.

Where b = take of
velocity.

d(t) =
-16t^2+bt+20.

d'(t) =
-32t+b.

So d'(t) = 0, when 32t + b = 0. Or t = b/32 = 32/32
= 1 second.

So with the initial velocity of 32 feet/sec,
the contestant attains the final velocity of zero ft/sec after one 1
second.

Since the bell is at 20 ft overhead, it is 20 feet
above the platform (or 40 above the ground).

d(1) = jump in
1 second, b = 32ft/sec. So we substitute the values in the equation: d(t)
=-16^2+32t+20.

d(1) = -16*1^2+32*1+20 = 36ft, which less
than 40ft.

So the contestant will not attain the height of
40ft with the initial velocity of 32 ft from the springboard. He can only attain a
maximum heght of 36 ft after which he falls.