Question

Y=-16x^2+190x+72 the height of the rocket y feet is related to the time after launch, x seconds, by the given equation

Answer 1

Answer: $636\ ft$

Step-by-step explanation:

Given

The height of the rocket is given by

$Y=-16x^2+190x+72$

Maximum height is given by the derivative of y and putting it to 0

$\Rightarrow \dfrac{dy}{dx}=-32x+190\\\\\Rightarrow -32x+190=0\\\Rightarrow x=\dfrac{190}{32}\\\Rightarrow x=5.93\ s$

Put the value of x in y

$\Rightarrow y=-16(5.93)^2+190(5.93)+72\\\Rightarrow y=-562.638+1126.7+72\\\Rightarrow y=636.06\ ft$