Ruby tosses a quarter off the main stree bridge into the st. john's river. the distance, in feet, the quarter is above the water is modulated by the equation d(t)= -16t^2+96t+112, where t represents time in seconds.
part a: from what height was the quarter tossed?
part b: how long does it take the quarter to reach its maximum height?
part c: what is the maximum height of the quarter?
part d: how much time does it take for the quarter to hit the water?

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Step-by-step explanation:

d(t) = -16t² + 96t + 112

A) The quarter's initial height is at t = 0.

d(0) = -16(0)² + 96(0) + 112

d(0) = 112

The quarter is tossed from a height of 112 feet.

B) The maximum height is at the vertex of the parabola, which is at x = -b/(2a).

t = -(96) / (2 × -16)

t = 3

The quarter reaches its maximum height after 3 seconds.

C) d(3) = -16(3)² + 96(3) + 112

d(3) = 256

The quarter reaches a maximum height of 256 feet.

D) When the quarter lands in the water, d(t) = 0.

0 = -16t² + 96t + 112

0 = t² − 6t − 7

0 = (t + 1) (t − 7)

t = -1 or 7

Since t can't be negative, t = 7.  So the quarter lands in the water after 7 seconds.

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17