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 +1#. What is the maximum height?

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Answer 1 · 2016-02-23T19:26:29

#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#.