Question

a football is thrown in the air.. the height, h(t), of the ball, in metres, after t seconds is modelled by this equation...

Answer 1

\[h(t) = -4.9(t-1.25)^{2} + 9\]

Answer 2

how high off the ground was the ball when it was thrown?

Answer 3

maximum height is 9 after 1.25 seconds

Answer 4

how did you find the maximum height?

Answer 5

oh oops

Answer 6

easy. the first part is a square. anything squared is positive. then when you multiply by -4.9 is will be negative, so the biggest the first part can be is 0. therefore the biggest the whole thing can be is 9

Answer 7

so, how high off the ground was the ball when it was thrown ?

Answer 8

plug in time

Answer 9

set derivative to 0 solve for t use t to find max height

Answer 10

answer to that is to expand and see what the constant is

Answer 11

you get 1.34375

Answer 12

like heck you do. you do not take the derivative and you do not set it to zero. you just see that the maximum is 9 because it is displayed for you

Answer 13

okay,
how high was the ball at 2.5 sec... (i got 1.3 metres)
and was the ball on its way up or down, how do you know?

Answer 14

maximum height is 9 yes?

Answer 15

my bad

Answer 16

and we know that it is 9 when x = 1.25 yes?

Answer 17

i believe soo..

Answer 18

which means that it is the highest after 1.25 seconds so after that it is heading down

Answer 19

goes up to 9 at 1.25 seconds, comes back down after

Answer 20

okay, makes sense now..
last question! when does the ball hit the ground?

Answer 21

set = 0 and solve for t, because when it hits the ground the height is 0

Answer 22

okay, thanks :]

Answer 23

\[-4.9(t-1.25)^2+9=0\]
\[(t-1.25)^2=\frac{9}{4.9}\]
\[x-1.25=\sqrt{\frac{9}{4.9}}\]
\[x=1.25+\sqrt{\frac{9}{4.9}}\]

Answer 24

whatever that is