l aballistossedintotheairat40feetpersecondfromaheightof5feet.howlongwill
Enter projectile motion problem

Solve the following projectile motion problem

aballistossedintotheairat40feetpersecondfromahof5feet.howlongwillittaketheballtoreachtheground

Projectile motion equation:

h(t) = -16t2 + vt + h where:
h = height, v = velocity, and t = time

Plug in our numbers

h(t) = -16t2 + (40)t + 5

When does the object hit the ground:

Set h(t) = 0

-16t2 + 40t + 5 = 0

Evaluate the quadratic

Only take positive values
since height cannot be negative

Quadratic Formula

  =  -b ± √b2 - 4ac
  2a

Calculate -b

-b = -(40)

-b = -40

Calculate the discriminant Δ

Δ = b2 - 4ac:

Δ = 402 - 4 x -16 x 5

Δ = 1600 - -320

Δ = 1920 <--- Discriminant

Since Δ > 0, we expect two real roots.

Take the square root of Δ

Δ = √(1920)

Δ = 8√30

-b + Δ:

Numerator 1 = -b + √Δ

Numerator 1 = -40 + 8√30

-b - Δ:

Numerator 2 = -b - √Δ

Numerator 2 = -40 - 8√30

Calculate 2a

Denominator = 2 * a

Denominator = 2 * -16

Denominator = -32

Find Solutions

Solution 1  =  Numerator 1
  Denominator

Solution 1  =  (-40 + 8√30)/-32

Analyzing the 3 terms in our 1st answer
We see that -40, -32, and 8 are all divisible by 8

Dividing them all by 8, we get: -5, -4, and 1

Solution 1 = (-5 . 1√30)/-4


Solution 2

Solution 2  =  Numerator 2
  Denominator

Solution 2 = (-40 - 8√30)/-32

Analyzing the 3 terms in our 1st answer
We see that -40, -32, and 8 are all divisible by 8

Dividing them all by 8, we get: -5, -4, and 1

Solution 2 = (-5 - 1√30)/-4


Solution Set

(Solution 1, Solution 2) = ((-5 . 1√30)/-4, (-5 - 1√30)/-4)


(Solution 1, Solution 2) = ((-5 . 1√30)/-4, (-5 - 1√30)/-4)