Question

The top of a building in Ottawa is 104m above the ground. Suppose an object was thrown upward with an initial velocity of 19.6m/s from this height. The approximate height of the object above the ground, h metres, t seconds after being thrown, would be given by the function:

h=-4.9t^2+19.6t+104

a) what was the maximum height of the object,above the ground?
b) after how many seconds did the object reach its maximum height?
c) From the time the object was initially thrown, how much did it takes to reach the ground, to the nearest second?

Answer 1

Do you just need the answers? If you need the work shown, are you doing differentiation and derivatives to do this or other methods?

Answer 2

i need the work shown and the ans please.

Answer 3

and i think were usin the quad formula

Answer 4

ohh, ok XD

Answer 5

x-coordinate of the Vertex or the maximum height is found using this equation: -b/2a
a=-4.9
b=19.6
-19.6/(2(-4.9))
=-19.6/-9.8
=2
x=2 for vertex
Plug it back in to figure out the y-coordinate

Answer 6

or t=2 in this case XD

Answer 7

h=-4.9(2)^2+19.6(2)+104
=-19.6+39.2+104
=123.6
(2, 123.6)

Answer 8

SO, the height is 123.6 and the time it took to get there is 2 seconds.

Answer 9

that's the answer for the first two questions

Answer 10

thanks :)
- from charmi too

Answer 11

welcome, now, for the third question right?

Answer 12

but like we learned in school , that the "b" value is the maximum height
thats the part we dont get :S

Answer 13

just a sec

Answer 14

I don't think that's right. The only way that would be true is if the coefficient of x^2 was 1/2.

Answer 15

ohh, so like the question isnt right? for part (c) ?

Answer 16

?? In question C, you just need to let h=0

Answer 17

ohh ok thanks :)