Aparticle is moving along a projectile path where the initial height is 160 feet with an initial speed of 144 feet per second. what is the maximum height of the particle? hint: use h(t) = -16t2 + vt + s. 128 feet 224 feet 272 feet 484 feet

D) 484 feet

Step-by-step explanation:

Given: , where t is the time, v is the initial velocity/speed and s is initial height.

and

v = 144 feet and s = 160 feet

Now plug in the given values in the given

This is in the form of quadratic function

The maximum height attain in vertex. Here we have formula to find the vertex of the x coordinate.

x =

Here b = 144 and a = -16

So, t =

The maximum height attain at t = 4.5 seconds. To find the maximum height plug in t = 4.5 in

H(4.5) =

= -16(20.25) + 648 + 160

= -324 + 648 + 160

= -324 + 808

= 484 feet

So the maximum height of the particle is 484 feet.

hello from mrbilldoesmath!

answer:

x - 3

discussion:

x^2 - 9 = (x-3) (x+3)   so

(x^2-9)/ (x + 3) = x-3

you,

mrb

because they could drop even lower..