SOLUTION: A rock is thrown upward from a bridge that is 83 feet above a road. The rock reaches its maximum height above the road 0.93 seconds after it is thrown and contacts the road 3.28 se
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Question 1148174: A rock is thrown upward from a bridge that is 83 feet above a road. The rock reaches its maximum height above the road 0.93 seconds after it is thrown and contacts the road 3.28 seconds after it was thrown.
Write a function
f
that determines the rock's height above the road (in feet) in terms of the number of seconds
t
since the rock was thrown. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A rock is thrown upward from a bridge that is 83 feet above a road.
The rock reaches its maximum height above the road 0.93 seconds after it is thrown and contacts the road 3.28 seconds after it was thrown.
Write a function f that determines the rock's height above the road (in feet) in terms of the number of seconds t since the rock was thrown.
:
-16t^2 + vt + 83 = h
find the velocity and height using:
t=3.28, h=0(road)
-16(3.28^2) + 3.28v + 83 = 0
-172.1344 + 3.28v + 83 = 0
3.28v - 89.134 = 0
3.28v = 89.13v
v = 27.175 ft/sec is the velocity
The equation
f(t) = -16t^2 + 27.175t + 83
:
find the height when t=.93
f(t) = -16(.93^2) + 27.175(.93) + 83
f(t) = -13.8384 + 25.7275 + 83
f(t) = 94.43 ft max height
:
green line 94.43
