can someone help me with this trigonometry question?!the height of a projectile can be modeled by the equation y-=16x^2+96x+256, where y is the...

This is not trogonometry. This is maths. This is dynamics
(, maths applied in physics.

We presume the equation of the
projectile y = -16x^2+96x+256.

So when x= 0, the height of the
projectile from the ground is y = 256.

If we put  y = 0 in the
equation we get  the time x the projectile reahes the ground.

dy/dx
= (-16x^2+96x+256) = -32x+96.

Whenx = 0, dy/dx = -16*0^2+96. So the
initial velocity u of the projectile = 96 ft/sec.

d2y/dx = g =
(-32x)' = -32 feet/sec^2 is the acceleration due to
gravity.

Therefore  -16x^2+96x+256 = 0  gives the solution for x
 when the projectile grounds.

-16x^2+96x+256 = 0 divided by -16,
gives x^2-6x+16 = 0.

x^2-6x+2 =
0.

(x+2)(x-8) = 0.

Therefore x= -2, or
x = 8 seconds.

So the projectile was projected 2 seconds ago from
the ground and now (when x= 0) it is at a height of y = -16*0^2+96*0+256 feet = 256 ft. So the
velocity at 2 seconds earlier should have been  u-2*32 = 160
ft/seconds.

So the projectile reaches its maximum height when its
vertical velocity is zero due to the downward ravitational pull. So   the final vertical
velocity  v = u-gx = o. Or 96-32x = 0. Or x = 96/32 = 3. So in another 3 seconds, the projectile
reaches the highest point given by : y(x) = -16x^2+96x+256 for the time x= 3seconds. So y(3) =
-16*3^2+96*3+256 =  400ft.

Velocity at the time of grounding v= 
u+gt = 96 -32*8 = - 160 ft/sec indicating that the direction of velocity is towards
ground.

Summary:

Status of the
projectile.

At time x = 0 (now), height of the projectile =
256.

The velocity of the projectile now(x= 0) = 96 ft/sec direction
is up.

Acceleration due to gravity = g = 32 feet per
second.

The velocity of the projectile t seconds ago when it had
started from the ground = 96 - (-32*2) = 160 ft/sec
upwards.

Maximumum height =
400ft.

Grounding time = after 8
seconds.

Grounding velocity = -160ft/sec  , direction
down.