If someone has a velocity of 32 ft/sec, will they be able to ring the bell( more info below)?At a carnival, a new attraction allows contestants to...

The initial velocity is u feet/sec. The bell is at a
height of 20 feet above the platform. Whether the contestant will be able to jump a 20
feet height from the spring board is the question. So the actual equation is h(t) =
ut-(1/2)gt^2 in accordance with the laws of motion, where h(t) height at time t from the
take off, and g is the acceleration due to gravity.

By
data  initial velocity u = by 32 ft/sec.  g = 32 ft/sec^2  a fact
assumed.

Therefore the the equation of motion is h(t) =
ut-(gt^2).

h(t) =
ut-(1/2)t^2.

So if the spring board has a height of 20 ft,
then the model would have to be h(t) =
32t-16t^2+20.

Therefore d(t) = -16t^2-bt+20 and
32t-16t^2+20 must be identical.

 When time t = 0, d(0) =
-16*0^2+32*0+20 = 20 is the platform height which is the initial height of the
contestant from where he takes off.

The bell is at 20 ft
above the platform. So the height of the bell = platform height + 20ft = 20ft +20ft = 40
ft.

By calculus, the maximum height d(t) is when d'(t) = 0
and d"(t) < 0.

d(t) =
ut-(1/2)gt^2+20 .

d'(t) = u-gt and d"(t) = -g <
0.

d'(t) = 0 gives u-gt= 0. So t = u/g , when
d(t) is maximum. Or d(u/g) is the maximum height the contestant can jump with an initial
velocity u.

When u =
32ft/sec:

d'(t) = -16*2t+32 and d"(t) = -32 <
0

So d'(t) = -16*2t+32 = 0, or  32t = 32. so t =
1

The maximum height  the contestant  jumps = d(1) =
-16*1^2+32+20 = 36 < 40ft.

Therefore the contestant
does not reach the bell.

If u = 35 ft/sec, then the maximum
height the contestant can jump  = d(u/g) = d(35/32) = -16(35/32)^2+35(35/32)+20 = 39.14
ft < 40 ft. So the contest  falls down before reaching the
bell.

If u = 40 ft/sec, then the contestant can jump a
maximum height of d(u/g) = -16(40/32)^2+40(40/32)+20 = 25+20 = 45 ft > 40 ft. So
the contestant  reaches a height above the bell. So he can ring twice the bell while
going up and falling down.

If u = 45 ft, then the maximum
height the contestant reaches d(t) = d(u/g) = d(45/32) = -16(45/32^2+45(45/32)+20 =
51.64 ft > 40 ft. So the contestant reaches a height above the bell. So he can
ring the bell twice. He can ring while going up and falling
off.

If u = 32 ft/sec, then the contestant  reaches a
maximum height of d(u/g) = d(32/32) = -16*1^2+32+20 = 36 feet = (20+16) ft. So the
contestant  jumps 16 ft above the platform. If the bell is placed at 10 ft, 12 ft and 15
ft above the platform, the contestant can reach and ring the bell. But he can not reach
the bell place at the  height of 18 feet as his maximum jump is only 16ft above the
platform.