Question

If a ball is thrown into the air with a velocity of 40 ft/s, its height (in feet) after t seconds is given by y = 40t − 16t^2. Find the velocity when t = 1.

help!

Answer 1

@zepdrix

Answer 2

the velocity is going to be equal to dy/dt

Answer 3

apply the derivative and then plug in t = 1

Answer 4

Don't know derivative yet :/

Answer 5

Unless..I knw to use
lim (h approaches 0)
[f(a+h)-f(a)]/h

Answer 6

ok we can use that

Answer 7

f(t) = 40t - 16t^2
f(t+h) = ???

Answer 8

f(t+h) = 40(1+h)-16(1+h)^2

Answer 9

use t instead of 1

Answer 10

well I guess you can plug in t = 1 later, so it works out

Answer 11

For now just keep it t?

Answer 12

I would keep t for now if I were doing it, but again you can plug in t = 1 when doing the limit definition

Answer 13

so

f(a+h) - f(a) = [ 40(1+h)-16(1+h)^2 ] - [ 24 ] = ??

Answer 14

40+40h-16+32h+16h^2-24

Answer 15

-16(1+h)^2 = -16(1+2h+h^2)
-16(1+h)^2 = -16-32h-16h^2

Answer 16

ooh dangit, that negative in front of the 16. Ah k.

Answer 17

Which simplifies to: 8h-16h^2

Answer 18

And further to: 8-16h
once we factor out an h and cancel it.

Answer 19

then plugging in 0 we get 8 :)

Answer 20

very nice, 8 is the correct answer

Answer 21

Thank you!

Answer 22

you're welcome