## anonymous one year ago The function H(t) = -16t2 + vt + s shows the height H(t), in feet, of a projectile launched vertically from s feet above the ground after t seconds. The initial speed of the projectile is v feet per second. Part A: The projectile was launched from a height of 96 feet with an initial velocity of 80 feet per second. Create an equation to find the time taken by the projectile to fall on the ground. Part B: What is the maximum height that the projectile will reach? Show your work. Part C: Another object moves in the air along the path of g(t) = 31 + 32.2t where g(t) is the height, in fe

1. anonymous

the object from the ground at time t seconds. Use a table to find the approximate solution to the equation H(t) = g(t), and explain what the solution represents in the context of the problem? [Use the function H(t) obtained in Part A, and estimate using integer values] (4 points) Part D: Do H(t) and g(t) intersect when the projectile is going up or down, and how do you know?

2. anonymous

@perl @perl

3. perl

Which part would you like help with?

4. anonymous

all of it.....

5. perl

6. anonymous

maybe once i get started i MIGHT be able to do it thanks so much

7. anonymous

H(t) = -16t^2 + 80t + 96

8. anonymous

i think ^^

9. perl

We are given that $$H(t) = -16t^2 + vt + s$$ shows the height H(t), in feet, of a projectile launched vertically from s feet above the ground after t seconds. The initial speed of the projectile is v feet per second. Part A) The projectile was launched from a height of 96 feet with an initial velocity of 80 feet per second. That means s = 96 and v = 80

10. perl

yes that's correct :)

11. anonymous

ok so my equation becomes : H(t) = -16t^2 + 80t + 96

12. anonymous

ok so thats part A right?

13. anonymous

now part b i have to find the maximum

14. anonymous

so the axis of symmetry is 2.5?

15. perl

the maximum or minimum of a parabola occurs at $$\large t= \frac { -b}{2a}$$

16. anonymous

ok so I got a maximum of 196

17. perl

how plug in that time t = 2.5 to find the maximum height

18. anonymous

196 :)

19. perl

correct

20. anonymous

ok so now part c... now i really dont know where to start

21. perl

Part C) the directions are not complete . Can you check

22. anonymous

yes hang on one sec please

23. anonymous

Part C: Part C: Another object moves in the air along the path of g(t) = 31 + 32.2t where g(t) is the height, in feet, of the object from the ground at time t seconds.

24. anonymous

Use a table to find the approximate solution to the equation H(t) = g(t), and explain what the solution represents in the context of the problem? [Use the function H(t) obtained in Part A, and estimate using integer values]

25. perl

We want to find when $$H(t) = g(t)$$ which would ordinarily mean solve the equation $$-16t^2 + 80t + 96= 31 + 32.2t$$ but instead of solving for t , here they ask us to make tables for the left side and right side separately, and estimate when they are closest

26. perl

|dw:1435527312454:dw|

27. anonymous

ok so I would solve each equation both H(t) and g(t) by plugging in the numbers 1-5?

28. perl

we can find y values for both H(t) and g(t) and then estimate where they are closest

29. anonymous

ok so we would do that by plugging in 0 right?

30. perl

yes

31. anonymous

ok so g(t) = 31 + 32.2t , g(t) = 31 + 32.2(0) , g(t) = 31 + 0 , g(t) = 31 H(t) = -16t^2 + 80t + 96 , H(t) = -16(0)^2 + 80(0) + 96 , H(t) = 0 + 0 + 96 , H(t) = 96

32. anonymous

so the y-intercept for H(t) = -16t^2 + 80t + 96 is 96 and the y-intercept for g(t) = 31 + 32.2t is 31

33. anonymous

@perl

34. anonymous

can you help me with part d @perl

35. anonymous

equation 1: (4,160) equation 2: (4, 159.8) what does it represent in the context of the problem?

36. perl

|dw:1435528016137:dw|

37. perl

|dw:1435528443369:dw|

38. anonymous

they almost intersect when x = 4 what is the meaning in the context of the problem?

39. anonymous

@perl

40. perl

at time t = 4 seconds the paths are closest to each other

41. perl

the projectiles are closest

42. anonymous

ok thanks! what about part d?

43. anonymous

when the projectiles re going down because H(t) = -16t^2 +80t +96 is decreasing just as the other equation is increasing?

44. anonymous

@perl

45. anonymous

@perl is that correct??

46. perl

when the projectile h(t) is coming down , yes

47. perl

the projectile g(t) is going up the whole time

48. anonymous

ok thanks so much !!!!!!

49. perl

|dw:1435528915392:dw|

50. perl

rough picture :)

51. anonymous

THANK YOU SO MUCH!

52. perl