Question

A toy rocket is launched vertically upward from a 12 foot platform with an
initial velocity of 128 feet per second. Its height h at timet seconds after
launch is given by the equation h(t) = -16t2 + 128t + 12. How many
seconds until the rocket is 40 feet?
0.23 seconds
4.5 seconds
5.67 seconds
7.77 seconds
8.09 seconds
11.34 seconds
12.52 seconds

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