# 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 jump off a springboard onto a platform to be launched vertically into the air. The object is to ring a bell located 20 ft overhead. The contestant's height in feet from the springboard is modeled by the function d(t)= -16t^2 -bt +20, where t is the time in seconds after leaving the platform, and b is the takeoff velocity from the platform.

If you could please show your work I don't understand this at all, so please explain in simple terms. Also if you want to be really nice and help me alot please answer these questions also, so one i learn how to do it i can check my answers. Im not trying to get you to do my homework i really need help and a high grade in math so pleasee help.

If someone has a velocity of 35 ft/sec, another person 40 ft/sec, and a third person 45 ft/sec will anyone of these people ring the bell.

Also, if bells were placed at heights of 10, 12, 15, and 18 ft off the ground, which ones will be reached with a velocity of 32 ft/sec.
I reallly need your help as i barely understand the question or if how to find the answer, so please help on at least one questionl. by the way this is a quadraticapplications word porblem.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.