# Need help on some algebra quadratic applicationsAt a carnival, a new attraction allows contestants to jump off a springboard onto a platform to be launched vetically into the air. The object is to...

Need help on some algebra quadratic applications

At a carnival, a new attraction allows contestants to jump off a springboard onto a platform to be launched vetically into the air. The object is to ring a bell located 20 ft overhead. The contestants 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 someone has a velocity of 32ft/sec, will they be able to ring the bell?

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### 1 Answer

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.