A ball is shot out of a cannon. The height of the ball, h in meters, as a function of the time, t in seconds, since it was shot is given by the equation $h = - 5 t^{2} + 30 t + 1$ $h= -5t^2 +30t +1$ . What is the maximum height?

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Answer 1 · 2016-02-23T19:26:29

$46 m$ $46m$

The maximum and minimum values will be at points where the derivative of this function is zero, ie
$EE h_(min//max)iff (dh)/dt=0$

$iff-10t+30=0$

$ifft=3s$ .

Since the second derivative $(d^2h)/(dt^2)|_(t=3)=-10<0$ ,
it implies that $t=3$ is a relative maximum of the function $h(t)$ .

$therefore h_(max)=h(3)$

$=-5(3^2)+30(3)+1$

$=46m$ .

A ball is shot out of a cannon. The height of the ball, h in meters, as a function of the time, t in seconds, since it was shot is given by the equation h= -5t^2 +30t +1h=−5t2+30t+1h= -5t^2 +30t +1. What is the maximum height?