It is possible to shoot an arrow at a speed as high as 80 m/s. If friction is neglected, how high would an arrow launched at this speed rise if shot straight up? How long would the arrow be in the air?
That is a typical arrow speed from a performance bow.
How high?
vtop=vinitial-g (time
time=80/9.8 seconds to the top, so double that to get total time in air.
Remember, you are neglecting air friction, which is considerable on those feathery things.
so how high does it go?
I cant figure it out
To determine the height the arrow would rise and the time it would be in the air, we can use the kinematic equations of motion. Let's break down the problem step by step:
Step 1: Find the time taken for the arrow to reach the highest point.
In order to find this, we need to determine the time it takes for the vertical velocity to decrease to zero. We can use the following equation:
v = u + at
Here,
v = final velocity (0 m/s at the highest point)
u = initial velocity (80 m/s)
a = acceleration due to gravity (-9.8 m/s², considering the arrow is moving straight up)
Substituting the values into the equation, we get:
0 = 80 - 9.8t
Solving for t:
9.8t = 80
t ≈ 8.16 seconds
Therefore, the arrow will take approximately 8.16 seconds to reach the highest point.
Step 2: Find the height the arrow will rise.
To calculate the height, we can use the equation:
s = ut + (1/2)at²
Here,
s = displacement (height)
u = initial velocity (80 m/s)
t = time (8.16 seconds)
a = acceleration due to gravity (-9.8 m/s²)
Substituting the values into the equation, we get:
s = (80)(8.16) + (1/2)(-9.8)(8.16)²
s ≈ 327.68 meters
Therefore, the arrow will rise to approximately 327.68 meters.
Step 3: Find the total time the arrow will be in the air.
Since the arrow goes up and then comes back down, the total time in the air is twice the time taken to reach the highest point:
Total time = 2 * t
Total time ≈ 2 * 8.16 seconds
Total time ≈ 16.32 seconds
Therefore, the arrow will be in the air for approximately 16.32 seconds.