Question

Three arrows are shot at different angles. A red arrow is shot at 30degrees, a blue arrow is shot at 45degrees, and a yellow arrow is shot at 60degrees.

A) Which arrow goes the farthest?
B) Which arrow goes the highest?
C) Which arrows go the same horizontal distance?
D) Which arrow is in the air for the shortest time?

Please help me out, let me know how to do it.

Answer 1

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.

Answer 2

ohh okay. thanks.

Answer 3

A. Range = Vo^2*sin(2A)/g. Let Vo = 10m/s for each arrow.
Range = 10^2*sin(60)/9.8 = 8.8m, Red.

Range = 10^2*sin(90)/9.8 = 10.2m, Blue.

Range = 10^2*sin(120)/9.8 = 8.8m, Yellow.

B. Let Vo = 10m/s for each arrow.
Yo = 10*sin30 = 5 m/s = Ver. component.
Y^2 = Yo^2 + 2g*h.
0 = 5^2 + (-19.6)h,
h = 1.28m, Red.

Yo = 10*sin45 = 7.07 m/s.
Y^2 = Yo^2 + 2g*h.
0 = 7.07^2 + (-19.6)h,
h = 2.55m, Blue.

Yo = 10*sin60 = 8.66 m/s.
Y^2 = Yo^2 + 2g*h.
0 = 8.66^2 + (-19.6)h,
h = 3.83m, Yellow.

C. Red and Yellow = 2.55m, each.

D. Xo = Vo*cos30 = 10*cos30 = 8.66 m/s = Hor. component.
Range = Xo*T.
8.8 = 8.66*T, T = 1.02s = Time in air, Red.

Xo = 10*cos45 = 7.07 m/s.
Range = Xo*T.
10.2 = 7.07*T, T = 1.44s, Blue.

Xo = 10*cos60 = 5 m/s.
Range = Xo*T.
8.8 = 5*T, T = 1.76s, Yellow.