Question

A toy rocket is launched vertically upward from a 12 foot platform with an

Answer 1

Answer:

The rocket is at 40 feet after 7.77 seconds.

Step-by-step explanation:

$h(t) = -16t^{2} + 128t + 12\\40= -16t^{2} + 128t + 12\\16t^{2} - 128t + 28 = 0\\4(4t^{2} - 32t + 7) = 0$

We use the quadratic formula to solve for t.

$t=\frac{32+\sqrt{32^{2}-4*4*7} }{2*4} \\t=\frac{32+\sqrt{1024-112} }{8}\\t=\frac{32+\sqrt{912} }{8}\\t=4+\frac{\sqrt{57} }{2}\\t=\frac{32-\sqrt{32^{2}-4*4*7} }{2*4} \\t=\frac{32-\sqrt{1024-112} }{8}\\t=\frac{32-\sqrt{912} }{8}\\t=4-\frac{\sqrt{57} }{2}$

We take the positive answer which is 7.77