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The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+4. The object is dropped again from a different starting height so that it hits the ground 2 seconds later. Write a function g that models the path of the object the second time it is dropped.

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Answer:

g(x) = - 16x² + 100

Step-by-step explanation:

Given the model:

First drop:

h(x) = - 16x² + 4

Second drop hits the ground 2 seconds later

Function g(x) to model the second drop:

-16t² + V0t + h0

h(x) = - 16x² + 4

0 = - 16x² + 4

16x² = 4

x² = 4/16

x² = 1/4

x = 1/2

2 seconds later :

X = 1/2 + 2

x = 2.5

0 = - 16(2.5)² + 0 + h0

0 = - 100 + h0

h0 = 100

Hence,

g(x) = - 16x² + 100

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