The function h(t) = -16^2 + 120t + 300 gives the height in feet of a projectile fired from the top of a building after t seconds. When will the object reach a height of 350ft?
If -16^2 + 120t + 300
mean
h(t) = -16 t² + 120 t + 300
then
-16 t² + 120 t + 300 = 350
subtract 350 to both sides
-16 t² + 120 t + 300 - 350 = 0
-16 t² + 120 t - 50 = 0
Multiply both sides by - 1
16 t² - 120 t + 50 = 0
Solve with the quadratic formula:
t1/2 = [ - b ± √ ( b² - 4 ∙ a ∙ c ) ] / 2 ∙ a
In this case a = 16 , b = - 120 , c = 50
t1/2 = { - ( - 120 ) ± √ [ ( - 120 )² - 4 ∙ 16 ∙ 50 ] } / 2 ∙ 16
t1/2 = [ 120 ± √ ( 14 400 - 3 200 ] / 32
t1/2 = ( 120 ± √11 200 ) / 32
t1/2 = [ ( 120 ± √ ( 1 600 ∙ 7 ) ] / 32
t1/2 = ( 120 ± √ 1 600 ∙ √7 ) / 32
t1/2 = ( 120 ± 40 ∙ √7 ) / 32
t1/2 = 40 ∙ ( 3 ± √7 ) / 32
t1/2 = 8 ∙ 5 ∙ ( 3 ± √7 ) / 8 ∙ 4
t1/2 = 5 / 4 ∙ ( 3 ± √7 )
t1 = 5 / 4 ∙ ( 3 - √7 ) = 5 / 4 ( 3 - 2.645751311 ) =
5 / 4 ∙ 0.354248689 = 0.44281086125 sec
t2 = 5 / 4 ∙ ( 3 + √7 ) = 5 / 4 ( 3 + 2.645751311 ) =
5 / 4 ∙ 5.645751311 = 7.05718913875 sec
if two piece of siopao cost 120 how many pieces of siopao can you buywith 300
To find when the object will reach a height of 350 feet, we can set up the equation:
350 = -16t^2 + 120t + 300
Let's solve for t:
-16t^2 + 120t + 300 - 350 = 0
Simplifying the equation:
-16t^2 + 120t - 50 = 0
Now, let's solve the quadratic equation. We can use the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
For our equation:
a = -16, b = 120, c = -50
Plugging the values into the quadratic formula:
t = (-120 ± √(120² - 4(-16)(-50))) / (2(-16))
Simplifying:
t = (-120 ± √(14400 - 3200)) / (-32)
t = (-120 ± √11200) / (-32)
t = (-120 ± 105.83) / (-32)
Now, we have two possible solutions:
t = (-120 + 105.83) / (-32) or t = (-120 - 105.83) / (-32)
Simplifying further:
t = (-14.17) / (-32) or t = (-225.83) / (-32)
t ≈ 0.44 or t ≈ 7.05
Therefore, the object will reach a height of 350 feet approximately 0.44 seconds and 7.05 seconds after being fired.
To find when the object will reach a height of 350 feet, we need to solve the given function for t when h(t) = 350.
The given function is h(t) = -16t^2 + 120t + 300.
Step 1: Set h(t) equal to 350 and solve for t:
350 = -16t^2 + 120t + 300
Step 2: Rearrange the equation to bring it into standard quadratic form (ax^2 + bx + c = 0):
-16t^2 + 120t + 300 - 350 = 0
-16t^2 + 120t - 50 = 0
Step 3: Divide the equation by -2 to simplify:
8t^2 - 60t + 25 = 0
Step 4: Solve the quadratic equation using factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values from our equation:
t = (-(-60) ± √((-60)^2 - 4(8)(25))) / (2(8))
Simplifying:
t = (60 ± √(3600 - 800)) / 16
t = (60 ± √2800) / 16
Step 5: Calculate the approximate values of t using a calculator:
t ≈ (60 ± 52.92) / 16
t ≈ (60 + 52.92) / 16 ≈ 7.88
t ≈ (60 - 52.92) / 16 ≈ 0.44
Therefore, the object will reach a height of 350 feet after approximately 0.44 seconds and 7.88 seconds.