A rock is thrown upward from a bridge that is 38 feet above a road. The rock rea

Question

A rock is thrown upward from a bridge that is 38 feet above a road. The rock reaches its maximum height above the road 0.64 seconds after it is thrown and contacts the road 3.68 seconds after it was thrown. Define a quadratic function, f, that gives the height of the rock (in feet) in terms of the number of seconds elapsed since the rock was thrown, t. A rock is thrown upward from a bridge that is 38 feet above a road. The rock reaches its maximum height above the road 0.64 seconds after it is thrown and contacts the road 3.68 seconds after it was thrown. Define a quadratic function, f, that gives the height of the rock (in feet) in terms of the number of seconds elapsed since the rock was thrown, t.

Explanation / Answer

since the rock is thrown upwards therefore quadratic equation is of the form

h(t)=-at^2 + bt +c

And initially the height is 38 feet

it means c=38

h(t)=-at^2 + bt + 38

and the maximum height is at t=0.64

It means vertex,t=0.64

And vertex=-b/2a

therefore,

-(-b)/2a=0.64

b=1.28a

And it hits the ground after 3.68 seconds

therefore

h(3.68)=0

0=-a(3.68)^2 + 3.68b+38

13.5424a-3.68b=38

now we have two equations that is

13.5424a-3.68b=38 and b=1.28a

on solving them we get

a=4.3 and b=5.5

Therefore the required equation is

h(t)=-4.3t^2 + 5.5t+38

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