PLEASE HELPPP URGENTTTT!!!!! The flight of a javelin released 6.5 feet..... ANSWER ALL PART PLZ

Question

The flight of a javelin released 6.5 feet from the ground at an initial velocity of 68 feet per second is modeled by the function h = -16t² + 68t + 6.5 where h is the height of the javelin in feet after t seconds.

Part 1:

Determine how many seconds it will take the javelin to reach the maximum height and find the maximum height. Round to the nearest hundredth if necessary

1) What are you being asked or told to do?

2) what do you know

3) what do we call the point where the maximum occurs

4) what is the axis of symmetry formula

5) What can you find using the axis of symmetry formula?

6) What is your plan for finding the maximum height

7) When does the javelin reach the maximum height? Show all work. Use the algebra skills you learned in Unit 4

8) What is the maximum height? Show all work. Use the algebra skills you learned in Unit 4

9) what does your answer mean

Part 2:

1) What do you know?

2) What is the height of the javelin when it hits the ground?

3) How can you adjust the given equation, above, to determine when the javelin will hit the ground?

4) what is the quadratic formula

5) Use the Quadratic Formula to calculate when the javelin hits the ground. Show all work

6) what does your answer mean

Answer 1

Part 1

3) Point where the maximum occurs is the vertex. Also it's the global and local or relative maximum. Graph the equation and you have a downward opening parabola, where the top point is the vertex.

4) axis of symmetry is t = 2 1/8 or t = 17/8 seconds, t =2.125

5) the axis of symmetry tells you the t or x coordinate of the vertex, the time when the javelin reaches maximum height

6) find the maximum height by taking the deriviative, h'(t) and setting it equal to zero, then solve for t. Then substitute that t value into the original h(t) equation h(2 1/8) = maximum height

7) time for maximum height is t= 2.125, same as the axis of symmetry

8) maximum height is h(2.125) = 78.75 feet

Part 2

1) from the equation which is a standard form, h(t) = (a/2)t^2 + vot + ho where ho = initial height = 6.5 feet, vo = initial velocity = 68 feet per second and a= the deceleration due to gravity, 32 feet per second per second. -16t^2 is actually rounded off to nearest integer.

2) height when it hits the ground is h= 0 feet

ground level means 0 feet off the ground.

3) h(t) = -16t^2 +68t +6.5

take the deriviative and set equal to zero to find maximum height

h'(t)=v(t) = velocity at time t = -32t +68 = 0

t =68/32 = 17/8 = 2.125 seconds or 2 1/8 seconds to reach maximum height or rounded off to 2.12 seconds

h(2.125) = -16(17/8)^2 +68(17/8) +6.5 = 78.75 feet = max height

(2.125, 78.75) is the vertex and maximum point on a downward opening parabola

h(2.125) = -16(17/8)^2 + 68(17/8) + 6.5 =-289/4 + 289/2 + 6.5 = 289/4 + 6.5 = 78.75

axis of symmetry is a vertical line through the vertex: t= exactly 2.125= 2 1/8 or round off to 2.13

4) and 5) it hits the ground when h(t)=0

-16t +68t +6.5 = 0

use the quadratic formula for the quadratic equation at^2 + bt +c, where a=-16, b= 68, c= 6.5

quadratic formula is: t=-b/2a + (1/2a)sqr(b^2-4ac)

ignore the negative square root

t =-68/-32 +(1/32)sqr(68^2+4(16)(6.5))

t=(68+71)/32

= about 4.34 seconds to reach the ground which is close to twice the time when it reahes maximum height. 2 1/8 x 2 = 4.25. It's slightly longer because it started off the ground, and that last 6.5 feet takes an additional 4.34-4.25 = 0.09 part of a second to reach the ground