A model rocket is launched with an initial velocity of 128 ft/s from a height of 70 ft. The height of the rocket, in feet, t seconds after it has been launched is given by the function s(t)=−16t^2+128t+70. Determine the time which the rocket reaches its maximum height and find the maximum height.
-16 t^2 + 128 t = s - 70
t^2 -8 t = -s/16 + 4.375
t^2 - 8 t + 16 = -s/16 + 20.375
(t-4)^2 = -(1/16) (s - 326)
vertex at time = 4 and height = 326
i dont get this bruh
To find the time at which the rocket reaches its maximum height, we need to determine the value of t at the vertex of the quadratic function s(t) = -16t^2 + 128t + 70.
The vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by the coordinates (h, k), where h represents the x-coordinate and k represents the y-coordinate of the vertex.
In our case, the quadratic function is s(t) = -16t^2 + 128t + 70. The formula for the x-coordinate of the vertex is given by h = -b/(2a), where a, b, and c are the coefficients of the function.
In this example, a = -16 and b = 128. Plugging in the values, we have:
h = -128/(2*-16) = -128/-32 = 4
Therefore, the rocket reaches its maximum height at t = 4 seconds.
To find the maximum height, we substitute the value of t into the function s(t):
s(4) = -16(4)^2 + 128(4) + 70
s(4) = -16(16) + 512 + 70
s(4) = -256 + 512 + 70
s(4) = 326
Therefore, the rocket reaches a maximum height of 326 feet.