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

a) 104 m

Answer 2

whats b

Answer 3

D: y is there 2 heights in the equation, theres like on (h) and then the max height (104)

Answer 4

idk thts wht the ques is
i found it frm our test paper

Answer 5

u have my test paper right -_-

Answer 6

yup

Answer 7

buh yu dint ans it

Answer 8

ohhh

Answer 9

u left it blank LOL

Answer 10

lmfaoo yee i dont know, ima try to do it

Answer 11

i posted it again
scroll up

Answer 12

im gunna post it