Question

can someone please

Answer 1

have u done parametric equations?

Answer 2

and do you know cartesian coordinates?

Answer 3

no:/

Answer 4

lets take the feet of the archer to be (0,0)

Answer 5

these are the points i got (0,1.39) (18,8) (45,0) :/

Answer 6

very good..now we need an equation which relates x and y dont we?

Answer 7

yeaah

Answer 8

lets say he launches the arrow with a horizontal velocity of v..the vertical velocity initaially is zero

Answer 9

can u tell me the x-coordinate at any point in time?

Answer 10

hmm, i don't know :/ ? im not very good at math.. ha

Answer 11

distance is speed x time..since there is no acceleration in the x-direction, the speed always remains v..so x = vt

Answer 12

understand?

Answer 13

yeah,, kindaa

Answer 14

similarly, there is a downward acceleration of -10 m/s^2 in the y-direction..
in the presence of accn, we have the distance covered as -0.5at^2, where a=accn

since we are starting at y=1.39 m, we remodel y as 1.39 - 0.5at^2, a= 10

therefore y= 1.39 - 5t^2

Answer 15

my teacher said that i would have to use quadratic equation to do this problem thou ..does this lead to that orr? :/

Answer 16

I posted an answer to this same question (but a different user posted it)

But the problem is I don't believe the height of A. You have it as 8 cm.
In the real world, this problem does not make sense. Are you sure about the height?

Answer 17

yeah, there's a diagram , i forgot what their called .. but next to the bulls eye it says A (8cm)

Answer 18

my teacher also said The distance from the top edge of the archery target to the ground is 1.5 meters

Answer 19

Here is the other post
http://openstudy.com/users/phi#/updates/507e201be4b0919a3cf31396

Answer 20

The distance from the top edge of the archery target to the ground is 1.5 meters

That is very important. Is the picture the same as in the post I just listed?

Answer 21

yess, it is

Answer 22

so you use the picture to find the height of A above the ground