Please if someone can help me with solving this problem..

Question

a projectile that is fired vertically into the air with an initial 120 ft. per second can be modeled by the equation s=120t-16t^2. In the equation , s is the distance in feet of the projectile above the ground after t seconds. The question is How long will it take for a projectile to reach 216 feet? and is it possible for the projectile to reach 900 feet? justify the answer. ...pls.

Answer 1

s = 120t - 16t²

 

How long denotes time. Therefore, we must find t when s=216.

 

216 = 120t - 16t²

 

-120t   -120t +16t²   

 

+16t²

 

 

16t² - 120t + 216 = 0

 

Divide by 8

 

2t² - 15t + 27 = 0

 

(2t - 9)(t - 3) = 0

 

t = 9/2   t =3

 

It will take 3 seconds for the projectile to be 216 feet above ground & it will come back to that same height in another 3/2 seconds.

 

Can the projectile reach 900 feet above ground?

 

16t² - 120t + 900 = 0

 

4t² - 30t + 225 = 0

 

When solving this quadratic equation, the answers are imaginary. Therefore the projectile cannot possibly reach 900 feet.

Answer 2

s= -16t²+120t

 

To know if thpe projectile can reach 900 ft we need to know the maximum height of the parabolic trajectory.  Since the square term is negative this will be a parabolic graph opening downward.  The maximum height reached will be the y value of the vertex.   in this case the horizontal axis is labelled t for time and the vertical acis is s for height.  To find the vertex. of a quadratic we use V=(-b/2a, f(-b/2a)) from the quadratic equation ...  s = at² + bt + c   a=-16, b=120, c=0

 

-b/2a=-120/2(-16)= 3.75

f(-b/2a) = -16(3.75)²+120(3.75)=225

The vertex is (3.75sec, 225 ft)

The projectile will reach a maximum height of 225 ft 3.75 seconds after launch.  It will not reach 900 ft.

 

How long will kt take to reach 216 feet?

 

-16t²+120t=216

-16t²+120t-216=0

This is a quadratic equation.

We can work with smaller numbers by factoring out -16.

-16(t² - 7.5t + 13.5)=0

t²-7.5t+13.5=0

a=1, b=-7.5, c=13.5

t= [-b ±√(b²-4ac)]/2a

t= [7.5 ±√((-7.5)²-4(1)(13.5))]/2

t= [7.5 ±√2.25]/2

t= (7.5 ± 1.5)/2

t = 9/2 = 4.5   or   t = 6/2 = 3

 

the projectile will reach a height of 216 feet 3 seconds after launch.

Continuing its trajectory it will peak at 225 ft as shown above, then again reach 216 ft 4.5 seconds after launch while on the downward arc of its trajectory.

Answer 3

s=120t-16t²

 

You are told that s=216 and your are to find t.

 

216=120t-16t²

 

This is a quadratic equation in t that can be solved for t.

 

16t²-120t+216=0

 

Using the formula for the solution of a quadratic equation gives:

 

t=3.75±0.75

 

The two solutions correspond to when the projectile is rising and then falling.  So it "reaches" 216 ft at

 

t=3.75-0.75=3 seconds.

 

To determine if the projectile reaches 900 feet, you need to know the vertex of the quadratic, where the projectile is at its highest.  This occurs at t=3.75 (the projectile follows an inverted parabola, so the peak is in the middle between 3.75-0.75 and 3.75+0.75, i.e., at t=3.75)

 

s=120(3.75)-16(3.75)²

 

s=225

 

The projectile's highest point is 225 feet, so it does not reach 900 feet.