# 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 height in feet and x is the time in seconds that the...

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 height in feet and x is the time in seconds that the projectile is in the air. find the greatest height of this projectile. find how many seconds it takes to attain this height. find the total time that the projectile is in the air.

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

y=-16x^2+96x+256

We need to determine the extreme values for the function y.

To do that , we will find the first derivative

==> y' = -32x + 96

Now let us calculate the critical values.

==> -32x + 96 = 0

==> -32x = -96

==> x =-96/ -32

==> x= 3

Then the function y has an extreme value when x= 0

==> y= -16(3^2) + 96*3 + 256

= -144 + 288 + 256

= 400

Then the extreme value is y= 300

Since the sign of x^2 is negative, then the function has a maximum value.

==> Then the greatest height is 300 and it takes 3 seconds to attain that height.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.

swavers | Student, Grade 11 | (Level 1) Honors

Posted on

sorry, the equation is y=-16x^2+96x+256 ****