Question

An archer shoots an arrow with an initial velocity of 21 m/s straight up from his bow. He quickly reloads and shoots another arrow in the same way 3.0 s later. At what time and height do the arrows meet?

Answer 1

https://in.answers.yahoo.com/question/index?qid=20101127235959AAy8Qtb

Answer 2

explain how to solve this pls

Answer 3

mmh okay

Answer 4

msg me once u r here

Answer 5

@aajugdar how do you solve this problem?

Answer 6

Vi = 21 m/s

Answer 7

first we have to find the max time to reach at maximum height for the first arrow

Answer 8

equation of motion is... u = v + at where u is final velocity

Answer 9

u = 0 v = 21 m/s a = -g ( while arrow moving upward).

Answer 10

so we have 0 = (21 m/s) - gt ===> g = 9.8 m/ sec^2

Answer 11

then what to do next?

Answer 12

so you have g= 9.8
put the value and you get time

Answer 13

which is 2.14

Answer 14

2.14 is the time when it reaches the top
so when its to the ground,then total time would be 4.28 s

Answer 15

now see
another arrow is thrown at same velocity

Answer 16

after 3 seconds

Answer 17

look at this

Answer 18

so it will meet first arrow when first arrow is coming down

Answer 19

okay, so what will be the time when they meet? and how do i solve for the height when the 2 arrows meet?

Answer 20

you will hv to find displacement

Answer 21

so you can see the formula for displacement in that link

Answer 22

s = ut +0.5at^2

Answer 23

u is initial velocity
which is 21
t is time which is 2.14

Answer 24

displacement of the arrow will be the height it reaches

Answer 25

now you know that 1st arrow is falling when 2nd is fired
right?

Answer 26

so it starts falling after 2.14 secs and 2nd arrow is fired after 3 seconds
so it will be falling for 3-2.14 = 0.86 secs when 2nd arrow is fired

Answer 27

So after falling for 0.85 seconds:
s = 0 + 0.5(9.8)(0.85)^2
= 3.54m
you understood this part?

Answer 28

its displacement from top is 3.54
so first arrow's height will be 22.5 - 3.54 = 18.960m

Answer 29

what about the 2nd arrow's height?

Answer 30

thats the height of first arrow when 2nd arrow is fired,so it will be 0 then

Answer 31

ok. so at what height do they meet?

Answer 32

before that you will have to find time
and before that velocity of first arrow when its coming down

Answer 33

Its velocity is:
v = at = 9.8(0.85)
=8.33m/s

Answer 34

now final part is easy
as the total distance between the two arrows just when arrow 2 is being fired is 18.96m, then the distance traveled by arrow 1 is 18.96-s, where s is the distance traveled by arrow 2.

Answer 35

now u know the equation
s= ut+0.5at^2

Answer 36

u hv velocities

Answer 37

all u hv to do is keep time same in both equations and put the values
(18.96-s) = 8.33t + 4.6t^2.....arrow 1
s = 21t - 4.6t^2.......arrow 2
subbing in arrow 1 into arrow 2 EQN yields:
18.96 - (21t - 4.6t^2) = 8.33t + 4.6t^2
29.33t = 18.96

Answer 38

t = 0.6464seconds This is the answer for time relative to the firing of arrow 2. relative to start of arrow 1's journey, time is 2.14 + 0.85 + 0.6464 = 3.636s

Answer 39

now you have time and velocity of first arrow
if u find dispacement of first arrow
it will be the height

Answer 40

the height where they meet

Answer 41

so
s = ut +0.5at^2
= 21(0.6464) - 4.6(0.6464)^2
= 11.65m from the ground

Answer 42

thats the height

Answer 43

ok. @aajugdar thank you so much!

Answer 44

yaw

Answer 45

My net connection was get off.... :(

Answer 46

@aajugdar shouldn't be the time 0.86 instead of 0.85?

Answer 47

yes
but wont make much of a difference