Tiffany kicks a soccer ball off the ground and in the air with an initial velocity of 28 feet per second. Using the formula H(t) = -16t2 + vt + s, what is the maximum height the soccer ball reaches

Question

anonymous · Answer

In this problem s=0 because the ball was initially on the ground. The maximum height is when H'(t) = 0.

anonymous · Answer

In other words, H'(t) = -32t + 33 = 0. Solving gives t = 33/32. Now plug into original equation and you will get the answer.

whpalmer4 · Answer

@mathgeekanqi Not sure how you came up with H'(t) = -32t+33 when v=28...
$$s=0,v=28$$$$H(t) = -16t^2+28t+0$$$H(t)$ is a parabola in form $y=ax^2+bx+c$ so the vertex is found at $$(x=-\frac{b}{2a}, y=ax^2+bx+c)$$Here we have $a=-16, b = 28, c = 0$ giving us $x=\large{-\frac{28}{2*(-16)} = \frac{28}{32} = \frac{7}{8}}$
Plugging $t=\large{\frac{7}{8}}$ into our formula for $H(t)$ we get the maximum height reached by the ball.  

No need for calculus, but I'll admit that I would go the calculus route if available rather than trust my memory for the vertex of a parabola!