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.
ohh okay. thanks.
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.
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.
To determine the answers to these questions, we will use some basic principles of projectile motion. In projectile motion, the distance and height traveled by an object depend on its initial velocity, angle of projection, and the acceleration due to gravity.
Let's analyze each question one by one:
A) To determine which arrow goes the farthest, we need to calculate the horizontal distances traveled by each arrow. The horizontal distance traveled by a projectile can be found using the formula:
Range = (Initial Velocity)^2 * sin(2 * Angle of Projection) / Acceleration due to Gravity
For the red arrow shot at 30 degrees, we need to know the initial velocity. Let's assume the initial velocity for all arrows is the same, say v. We'll substitute these values into the formula:
Range of Red Arrow = v^2 * sin(2 * 30) / g
Range of Blue Arrow = v^2 * sin(2 * 45) / g
Range of Yellow Arrow = v^2 * sin(2 * 60) / g
Comparing these three ranges, you can determine which arrow goes the farthest.
B) To determine which arrow goes the highest, we need to calculate the vertical distances (heights) reached by each arrow. The maximum height reached by a projectile can be determined using the formula:
Maximum Height = (Initial Velocity)^2 * sin^2(Angle of Projection) / (2 * Acceleration due to Gravity)
Using the same assumptions and calculations, you can find the maximum height reached by each arrow by substituting the values into the formula provided. Comparing these heights will give you the answer.
C) To determine which arrows go the same horizontal distance, you need to compare the ranges. From the previous calculations, you can compare the values obtained for the range of each arrow. If two or more arrows have the same range, they will cover the same horizontal distance.
D) To determine which arrow is in the air for the shortest time, we need to calculate the time of flight for each arrow. The time of flight is the total time the arrow spends in the air. It can be calculated using the formula:
Time of Flight = 2 * Initial Velocity * sin(Angle of Projection) / Acceleration due to Gravity
Again, by substituting values into the formula for each arrow, you can compare the times of flight and determine the one that stays in the air for the shortest duration.
Keep in mind that we made some assumptions, such as the initial velocity being the same for all arrows and neglecting other factors like air resistance. These assumptions may affect the accuracy of the results, but they should provide a good estimation.