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
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
5 answers
sorry i still dont understand how to calculate it ?
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
thanks very much
You are welcome.