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Mathematics 15 Online
OpenStudy (anonymous):

A ball is thrown into the air with an upward velocity of 40 ft/s. Its height h in feet after t seconds is given by the function h=-16t^2+40t+6. A. In how many seconds does the ball reach its maximum height? B. What's the ball's maximum height?

OpenStudy (bahrom7893):

A) h'(t) = -32t + 40 = 0 -32t = -40 t = 5/4

OpenStudy (bahrom7893):

Test if it's max or min: h" = -32 < 0 therefore it's max

OpenStudy (bahrom7893):

B) The ball's maximum height therefore occurs at t = 5/4 and the height is: h(5/4) = -16(5/4)^2 + 40(5/4) +6

OpenStudy (anonymous):

wait i dont get what you are doing in the first slide

OpenStudy (bahrom7893):

Taking the derivative of the function.

OpenStudy (bahrom7893):

Did you take calculus?

OpenStudy (anonymous):

Yah but you cant square the 32 yet you have to find the t first

OpenStudy (anonymous):

no this is Algerbra advance

OpenStudy (bahrom7893):

sorry nanas.. I am doing everything correct, but I don't know how to do this using algebra. Wait actually if u use algebra, just plot the graph.. look at it's highest point, x-value will be the time and y-value will be the height.. This is a parabola pointing down..

OpenStudy (bahrom7893):

I was using calculus and my method works too, but if u don't know it i can just give u answers..

OpenStudy (anonymous):

I tried that i guess my brain is to advanced lol no I dont get how to plo the points but the anwser would be fine

OpenStudy (anonymous):

i mean how to plot

OpenStudy (bahrom7893):

just type in the equation and then use trace.. well the answers would be: A) 5/4 B) -16 * (5/4) ^2 + 40 * (5/4) + 6 <=whatever that equals.. sorry don't feel like doin arithmetic right now.. i actually gotta go to bed lol got a class early 2morro..

OpenStudy (vijay):

t = 10 secs

OpenStudy (anonymous):

wow what time is it there

OpenStudy (bahrom7893):

vijay can't be..

OpenStudy (bahrom7893):

10:30 pm

OpenStudy (vijay):

02:30 A.M.

OpenStudy (anonymous):

wow its only 7 30 here

OpenStudy (bahrom7893):

vijay t = 10 secs doesn't make sense.. it's t = 5/4

OpenStudy (vijay):

if u use equation v=u+ft v=0, u=40, f= -10, t= ?

OpenStudy (anonymous):

ok What is the lil star thing

OpenStudy (bahrom7893):

he is asking maximum height..

OpenStudy (bahrom7893):

little star is multiplication

OpenStudy (bahrom7893):

time at max height..

OpenStudy (vijay):

1st find the time, then plug the t into the function, u will get height

OpenStudy (anonymous):

bahrom is right i remember my online teacher saying this

OpenStudy (anonymous):

oh so all i got to do is the multiplication

OpenStudy (bahrom7893):

Okay here's proof: At t = 10; h = -16*(10)^2+40*10+6 = -1600 + 400 + 6 is less than 0.. u can't have a negative maximum height.. the ball is still in the air..

OpenStudy (bahrom7893):

yeah just plug in 5/4 for t in the h(t) equation and find the answer..

OpenStudy (bahrom7893):

Please click on become a fan if I helped, I really want to get to the next level!! Thanks =) I'm off..

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