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anonymous
 one year ago
A rock is launched from a cannon. Its height, h(x), can be represented by a quadratic function in terms of time, x, in seconds.
After 1 second, the rock is 129 feet in the air; after 2 seconds, it is 236 feet in the air.
Find the height, in feet, of the rock after 5 seconds in the air.
anonymous
 one year ago
A rock is launched from a cannon. Its height, h(x), can be represented by a quadratic function in terms of time, x, in seconds. After 1 second, the rock is 129 feet in the air; after 2 seconds, it is 236 feet in the air. Find the height, in feet, of the rock after 5 seconds in the air.

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welshfella
 one year ago
Best ResponseYou've already chosen the best response.0we can use the established equation of motion under acceleration due to gravity( = 32 ft s2 ) and the equation is s = ut + 0.5at^2 where u = initial velocity , a = acceleration = 32 , s = distance and t = time s = ut  16t^2 so plugging in given values 129 = u(1)  16(1)^2 129 = u  16 u = 145 so our equation is s = 145t  16t^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so whats the answer? haha sorry im really bad at thinking right now

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0when t = 5 height s = 145*5  16*5^2 = 325 feet

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0h(x) after 5 seconds is 325 feet

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0well thats the correct equation

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0another value of u is 150 which will give a result of 350

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0if i use the 236 ft after 2 seconds it gives u = 150
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