Question

An archer of the Philippine Team practices daily in arrow shooting. One time she experimented shooting vertically upward with a velocity of 29.4 m/s. One second later, she shoots another arrow with the same speed and direction as the first arrow. Will the two arrows meet? If so, where and when? (Kindly show me the formulas and solutions. I'm kinda catching up with my class, thanks :))

Answer 1

so in the question no drag or something else?we must know when the first arrow stops moving !

Answer 2

Air resistance is ignored.

Answer 3

Topic is about motion along a straight line. Acceleration due to gravity is -9.8 m/s

Answer 4

well it's ignored...

Answer 5

i do not know what formulas to use but i think based on class that v initial= 29.4 m/s

Answer 6

so if we want to know when the first arrow stops moving then you mean its the highest point before it goes downward again right?

Answer 7

yes

Answer 8

Vf= Vi - g(t) can be used ?

Answer 9

Vf= 29.4 - 9.8(t)
What value (time) should i use?

Answer 10

well i think this would help you...i don't know ! maybe i cannot help you in this case ;)

Answer 11

its ok. thanks for trying :))

Answer 12

First we calculate at what time and height arrow1 stops in mid air.
Than we use this information to make new equations for arrow 1 and 2 from their positions and speeds at the time the first arrow stops in mid air to see where they meet

arrow 1 : \[v = 29.4 - 9.8t\]
gives\[t = 3s\]
the arrow reaches a height of :
\[x = 29.4 - 9.8*3 ^{2}/2 = 44.1m\]
at this time arrow 2 has reached :
\[x _{2} = 29,4 - 9.8(3-1)^{2}/2 = 39.2m\]
the speed of arrow2 is now :
\[v _{2}=29.4-9.8\times2=9.4m/s\]
equations for arrow 1 and 2 are now :
\[x _{1}=44.1-9.8t ^{2}/2\]
\[x _{2}=39.2+9.8t-9.8t ^{2}/2\]
when they meet \[x _{1}=x _{2}\]
so : \[44.1-9.8t ^{2}/2=39.2+9.8t-9.8t ^{2}/2\]
which yields : \[t=0.5s\]

Answer 13

The two arrows meet at : \[x=39.1+9.8*0.5-9.8*0.5^{2}/2=42.775m\]

Answer 14

thanks, ill study your calculations in further studies

Answer 15

you're welcome.