Posted by Anonymous on Tuesday, November 18, 2008 at 9:23pm.
First, the time in air:
h=visin45*time -1/2 g time^2
so when does the arrow return to base height?
0=visin45*time-1/2 g t^2
0=(visin45-1/2 g t)t
or time in air is 2*vi*sin45/g
height: when is vv=0?
vv=visin45-gt which is zero at
time=vi*sin45/g
hmax=vi*sin45*vi*sin45/g-1/2 g (visin45/g)^2 or
hmax=1/2 (vi*sin45/g)^2
and so on. for the other angles, put those numbers in and see, the only variable is the sine of the angle.
ohh okay. thanks.
Related Questions
physics - Two arrows are shot vertically upward. The second arrow is shot after ...
Physics - Two arrows are shot vertically upward. The second arrow is shot after ...
physics - Two arrows are shot vertically upward. The second arrow is shot after ...
physics - An arrow is shot verticaaly upward. Two seconds later, it is at a ...
Physics - An arrow is shot into the air at a velocity of 25 m/s and an angle of ...
physics - You are watching an archery tournament when you start wondering how ...
Physics - An arrow is shot at an angle of above the horizontal. The arrow hits a...
Physics - An arrow is shot at 29.0° above the horizontal. Its initial speed...
Physics - An arrow is shot at 29.0° above the horizontal. Its initial speed ...
Physics - An arrow is shot a 30.0° above the horizontal. Its velocity is 49 ...
For Further Reading