Question

A ball is thrown vertically upward with an initial velocity of 80 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s=80t-16t^2.
a) at what time t will the ball stick the ground?
b) for what time t is the ball more then 64 feet above the ground?

Answer 1

when the ball hits the ground s=0

Answer 2

80t-16t^2=0

Answer 3

do u know how to solve that equation?

Answer 4

do u put it in the quadratic equation??

Answer 5

\[16t(5-t)=0\]

Answer 6

u take the common factor...

Answer 7

so either t=0 or t=5

Answer 8

\[\therefore t=5\sec\]

Answer 9

so that is the answer for part (a)

Answer 10

if u got that one so i can move to solve part b

Answer 11

okay

Answer 12

did u understand the first part?

Answer 13

u don't sound like u did

Answer 14

yeah u do 16t = 0 and (5-t)=0 then find for t

Answer 15

ok good, so i did not choose t=0 to be the correct timing why?

Answer 16

anyways let's do part b where u were asked to find the time at s=64ft

Answer 17

64=80t-16t^2.

Answer 18

do u know how to solve this one?

Answer 19

16t^2-80t+64=0

Answer 20

divide both sides by 16

Answer 21

are u there?

Answer 22

not sure why you would divide 16

Answer 23

to simplify the solution

Answer 24

because u always want the coefficient of t^2 to be equal to 1

Answer 25

okay but how will u divide 0 from 16

Answer 26

16t^2-80t+64=0 ->t^2-5t+4=0

Answer 27

0/16=0

Answer 28

take the common factor which ever way u want to look at it all the roads lead to Rome...

Answer 29

either t=1 or t=4

Answer 30

okay. so the answer would be 1<x<4

Answer 31

you mean t

Answer 32

yes t!

Answer 33

yes

Answer 34

if t=0 that means the maximum height is 64ft

Answer 35

did u get the answer?

Answer 36

so the answer would be 1<t< 4 right for b

Answer 37

i replied to u but it seems my msg was not posted, yes you are right