An Arrow directed at a stationary target is shot at an angle of E25N. The arrow hits the intended target at the same height from which it is launched, 3.0s later.
Determine the distance from the archer to the target.
2 answers
It matters how fast the arrow was launched, and at what elevation.
E25N ???
are you sure you do not mean 25 degrees above horizontal??
I will assume that is what you mean
S = initial speed
constant horizontal speed = u = S cos 25
initial up speed = Vi = S sin 25
distance d = u t = 3 S cos 25
time in air = t
hits max at t/2
0 = Vi - g (t/2) = Vi - 4.9 t in metric units
Vi = 4.9*3 = 14.7 meters/second
so
S sin 25 = 14.7
solve for S
u = S cos 25
solve for u
d = 3 u the end
are you sure you do not mean 25 degrees above horizontal??
I will assume that is what you mean
S = initial speed
constant horizontal speed = u = S cos 25
initial up speed = Vi = S sin 25
distance d = u t = 3 S cos 25
time in air = t
hits max at t/2
0 = Vi - g (t/2) = Vi - 4.9 t in metric units
Vi = 4.9*3 = 14.7 meters/second
so
S sin 25 = 14.7
solve for S
u = S cos 25
solve for u
d = 3 u the end
1 answer