## anonymous one year ago A ball is thrown vertically upward from the top of a 200 foot tower, with an initial velocity of 5 ft/sec. Its position function is s(t) = –16t2 + 5t + 200. What is its velocity in ft/sec when t = 3 seconds?

1. anonymous

2. anonymous

3. anonymous

@robtobey

4. anonymous

take the derivative and plug in 3 for t

5. anonymous

is that how you find the distance? @sourwing

6. anonymous

no. velocity

7. anonymous

WHT DO YOU MEAN BY DERIVATE?

8. anonymous

@sourwing ?

9. anonymous

10. anonymous

@chrisdbest

11. Kash_TheSmartGuy

Answer to a similar question - http://www.algebralab.org/Word/Word.aspx?file=Algebra_MaxMinProjectiles.xml

12. anonymous

they're similar but they talk about the maximums @Kash_TheSmartGuy

13. Kash_TheSmartGuy

Oh ok.

14. anonymous

do you know someone that could help me? @Kash_TheSmartGuy

15. Kash_TheSmartGuy

Probably @jim_thompson5910 or @campbell_st

16. anonymous

@jim_thompson5910 could you help me?

17. jim_thompson5910

this is for calculus class right?

18. anonymous

yes

19. jim_thompson5910

have you learned about derivatives? or differentiation?

20. anonymous

No I haven't, but my professor explained that it was related to the slope formula

21. jim_thompson5910

that's the average velocity, but they want the instantaneous velocity here

22. jim_thompson5910

23. anonymous

Wait I checked my notes isn't f(x) lim x-->0 f(a-h)-f(a) / h

24. jim_thompson5910

yes it is

25. jim_thompson5910

that's the limit definition of the derivative

26. anonymous

okay, but what could I first?

27. anonymous

Do I plug the 3 in the A right

28. jim_thompson5910

yes

29. jim_thompson5910

what are f(3) and f(3+h) equal to

30. jim_thompson5910

btw that should be f(a+h) and not f(a-h) but I guess it doesn't matter if h goes to 0

31. anonymous

F(3) gives me 71 and F(a+h) gives me h^2+6h+9

32. jim_thompson5910

f(3) is correct but what you got for f(3+h) is not correct

33. anonymous

isn't -144-96h-16h^2

34. jim_thompson5910

f(t) = -16t^2+5t+200 f(3+h) = -16(3+h)^2+5(3+h)+200 f(3+h) = -16(9+6h + h^2)+5(3+h)+200 f(3+h) = -144-96h-16h^2+15+5h+200 f(3+h) = -16h^2-91h+71

35. jim_thompson5910

Use f(3) = 71 f(3+h) = -16h^2-91h+71 to find $\Large \frac{f(3+h)-f(3)}{h}$

36. anonymous

okay i see my mistake -91

37. jim_thompson5910

yep $\Large \frac{f(3+h)-f(3)}{h}=\frac{-16h^2-91h+71-71}{h}$ $\Large \frac{f(3+h)-f(3)}{h}=\frac{-16h^2-91h}{h}$ $\Large \frac{f(3+h)-f(3)}{h}=\frac{h(-16h-91)}{h}$ $\Large \frac{f(3+h)-f(3)}{h}=\frac{\cancel{h}(-16h-91)}{\cancel{h}}$ Then plug in h = 0 to get -16h-91 to turn into -91

38. jim_thompson5910

at that instant in time, the object is going -91 ft/sec (ie it's falling at 91 ft/sec)

39. anonymous

okay, OMG thank you very muchI will write this down about the instantaneous velocity formula. I really appreciate all your help, god will pay you in a way that you will feel blessed. thank you, I had to submit this today at 12 and I am ready, thank you so much

40. jim_thompson5910

you're welcome, I'm glad it helped out