+80 A ball travels on a parabolic path in which the height (in feet) is given by the expression -16t^2+64t+31, where t is the time after launch. What is the maximum height of the ball, in feet?
+6251 \(\text{can you use calculus?}\\ \text{If so we can find the first derivative of the height (it's vertical velocity)}\\ \text{and solve the the }t \text{ that makes this zero}\\ \text{(the ball stops for a moment at the top of it's arc)}\\ v_z(t) = \dfrac{d}{dt} h(t) = -32t + 64\\ v_z(t) = 0 \Rightarrow t = 2\\ h(2) = -16(4) + 64(2) + 31 = 95\)
\(\text{If you can't use calculus (and I suspect you can't)}\\ \text{we can rewrite }h(t) \text{ in standard parabola form}\\ -16t^2 + 64t + 31 = \\ -16(t^2 -4t) + 31 = \\ -16((t-2)^2-4)+31 = \\ -16(t-2)^2 + 64 + 31 = \\ -16(t-2)^2 + 95\\ \text{and it should be clear this reaches a maximum of }\\ h=95 \text { at }t=2\)
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+37146 Max height will be at time, t = -a/(2b) = - 64/(2*-16) = 2 seconds
Substitue this time into the equation to find max height
-16(2^2) + 64(2)+31 = 95 ft max
