Question

An arrow is shot by an archer at a 50.0 degree angle at a velocity of 10.0 m/s. What is the maximum height and range of the arrow?

**how do you calculate this? thank you:)

Answer 1

we want the hieght when the final velocity in the y direction is 0

Answer 2

yes:)

Answer 3

so Vf=Vi+at

Answer 4

but we are dealing with the y direction so

Answer 5

okay:)

Answer 6

10.0sin(50) = initial velocity in the Y direction

Answer 7

the acceleration is from gravity so its negative which is a constant -9.8m/s^2

Answer 8

well actualy i think we can use a different equation for htis to make it easier

Answer 9

okie:)

Answer 10

Vf^2=Vi^2+2ay i think it is

Answer 11

0=(10sin(50))^2-2(9.8)(Y)

Answer 12

solve for Y

Answer 13

-[10sin(50)]^2=-2(9.8)(Y)

Answer 14

10sin(50)/(2)(9.8) = Y

Answer 15

okay, so Y=-12.856?

Answer 16

lul wait wait

Answer 17

haha okie

Answer 18

oh i see

Answer 19

i forgot to square it one sec

Answer 20

ok!

Answer 21

im getting 3m

Answer 22

http://www.wolframalpha.com/input/?i=%2810sin%2850%29%29^2%2F%282%289.8%29%29

Answer 23

ohh okay! so this is the max height correct? 2.99=3 is the maximum height?

Answer 24

ya

Answer 25

awesome!! how do we find the range now?

Answer 26

u mean where the arrow lands?

Answer 27

in this case we need to find time.

Answer 28

yes:)

Answer 29

how do we do that?

Answer 30

so 10cos(50) = velocity in the x directiion

Answer 31

ah i see so we need to find when the ball hits the ground

Answer 32

to do that we have to consider whats happening in the Y direction first

Answer 33

okie:)

Answer 34

we already solved its max height

Answer 35

now the time it hits the ground is equal to 2 times the time it takes for it to reach the max height

Answer 36

yes:) 3.00!

Answer 37

so we use the kinematic equation

y=Vi(t)+1/2at^2

Answer 38

or actually better to use

Answer 39

Vf=Vi+at

Answer 40

okay! so what do we plug in?

Answer 41

so 0=10sin(50)-9.8(t)

Answer 42

-10sin(50)=-9.8t

Answer 43

10sin(50)/9.8 = t

Answer 44

however this t corresponds to the time when the arrow is at its max height

Answer 45

and we want the time that is twice that

Answer 46

ahh okay, and the t=-0.2677?

Answer 47

t = 1.56

Answer 48

well the t in the equation is half that

Answer 49

but we multiply it by 2 to get the time for it to hit the ground rather then the time to reach its max height

Answer 50

so then we have x=Vi(t)

Answer 51

ohh okay!

Answer 52

x=10cos(50)(1.56)

Answer 53

should give u ure answer

Answer 54

ahh okie:) so

x=15.053

that is the range?

Answer 55

x=10

10cos(50)(1.56)

Answer 56

http://www.wolframalpha.com/input/?i=10cos%2850%29%281.56%29

Answer 57

ohh okay! yay!!
so mx height=3 and range=10 ?

Answer 58

ya

Answer 59

yay! thank you !! :)