Question
An archer shoots at an angle of 11 degrees from the horizontal at an initial speed of 31.8m/s. Ignoring air resistance, determine
a) the time it will take for the arrow to strike the ground
b) the range of the arrow
a) the time it will take for the arrow to strike the ground
b) the range of the arrow
Answers
Damon
vertical problem first
Vi = 31.8 sin 11
v = Vi - 9.81 t
at top v = 0
t at top = (31.8 sin 11) / 9.81
t in air = 2 * t at top = 2 (31.8 sin 11) / 9.81 because parabola symmetric )
then horizontal problem now we know how long in air
U = horizontal speed = 31.8 cos 11
There is NO horizontal force
SO there is no change in horizontal speed (until it hits ground)
so
we know t in air and U
distance = U t
Vi = 31.8 sin 11
v = Vi - 9.81 t
at top v = 0
t at top = (31.8 sin 11) / 9.81
t in air = 2 * t at top = 2 (31.8 sin 11) / 9.81 because parabola symmetric )
then horizontal problem now we know how long in air
U = horizontal speed = 31.8 cos 11
There is NO horizontal force
SO there is no change in horizontal speed (until it hits ground)
so
we know t in air and U
distance = U t
Zoe
sorry i still dont understand how to calculate it ?
Damon
t in air = 2 * t at top = 2 (31.8 sin 11) / 9.81
and
distance = 31.8 cos 11 * t in air
and
distance = 31.8 cos 11 * t in air
Zoe
thanks very much
Damon
You are welcome.