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

The rocket is at 40 feet after 7.77 seconds.

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=(32+√(1024-112) )/(8)\\t=(32+√(912) )/(8)\\t=4+(√(57) )/(2)\\t=\frac{32-\sqrt{32^(2)-4*4*7} }{2*4} \\t=(32-√(1024-112) )/(8)\\t=(32-√(912) )/(8)\\t=4-(√(57) )/(2)`

We take the positive answer which is 7.77