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Applied Mechanics , Please explain step by step!!?
1) An arrow is shot straight up at 40 m/s, and 2 seconds later a second arrow is shot up at 60 m/s.
a)Where do the arrows cross?
b)When do the arrows cross?
1 Answer
- Andrew SmithLv 74 years agoFavorite Answer
I saw this one yesterday. It is a mathematical exercise not a physics one.
You prepare two simultaneous equations and solve.
1. s = 40 t - g t^2
2. s= 60 (t-2) - g( t-2)^2
you must expand out 2.
s = 60 t - 120 - gt^2 + 4gt - 4g
= -gt^2 +(60+4g) t - (120+4g)
now equate with 1
-gt^2 +(60+4g) t - (120+4g)= 20t - gt^2
add gt^2 to both sides
(60 + 4g)t - (120+4g) = 20t
(40 + 4g) t = 120 + 4g
t = (120+4g) / ( 40 + 4g)
Substitute your value of g.
To find t then substitute back to find s.
