Question

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.

Answer 1

we can use the established equation of motion under acceleration due to gravity( = 32 ft s-2 ) 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

Answer 2

so whats the answer? haha sorry im really bad at thinking right now

Answer 3

when t = 5
height s = 145*5 - 16*5^2 = 325 feet

Answer 4

it was incorect

Answer 5

h(x) after 5 seconds is 325 feet

Answer 6

well thats the correct equation

Answer 7

another value of u is 150 which will give a result of 350

Answer 8

if i use the 236 ft after 2 seconds it gives u = 150