The equation is y=-16x^2+110x+4

Question

From what  height was the object launched?

 

What was the height of the object  3 seconds into its flight?

 

How long  will  the object  stay in the air before landing 

Answer 1

Hello Maria,

 

From what height was the object launched?

This is when x = 0, so solve the equation for x=0 and the first two terms will reduce to 0, leaving y=4.  

The launching height is 4.

 

What was the height of the object 3 seconds into its flight?

This is when x = 3.

y = -16(3²) + 110(3) + 4

y  = -144+330+4

y = 190

The height at 3 seconds is 190.

 

How long will the object stay in the air before landing.

Here you want to know when the object reaches the ground, i.e. when will y=0.

So,

0=-16x²+110x+4

Let's factor out a -2

0 = (-2) (8x²-55x-2)

and

0 = (8x²-55x-2)

 

Let's use the quadratic formula to solve for x.

 

-b ± √(b²-4ac)       -(-55) ± √[(-55)2-(4)(8)(-2)]

-----------------  =  ----------------------------------

       2a                                  2(8)

 

Let's clean this up.

 

55 ± √(3025+64)      55 ± √(3089)

-------------------- =  ----------------- 

         16                           16

 

Since 3089 is a prime number, we can only approximate it as 55.58

 

x = 55 ± 55.58        110.58          -0.58

      -------------  =  ---------- or   -------

           16                   16              16

 

This second response is not possible as we cannot have a negative solution.

110.58

--------  ≡  6.9

   16

The object will stay in the air approximately 6.9 seconds.