An arrow is aimed horizontally, directly at a target 20.6 meters away. when the arrow is released, it lands 0.095 meters below the target point.
a) How long was the arrow in the air?
b) How fast was the arrow moving when it left the bow?
a. d = 0.5g*t^2 = 0.095 m.
4.9t^2 = 0.095
t^2 = 0.0194
Tf = 0.139 s. = Fall time or time in air.
b. Dx = Xo*Tf = 20.6
Xo * 0.139 = 20.6
Xo = 148 m/s.
To find the answers to these questions, we can use the principles of projectile motion. In this case, since the arrow is being launched horizontally, we can just consider the vertical motion.
a) To determine the time the arrow was in the air, we need to find the time it takes for the arrow to fall 0.095 meters. We can use the formula for vertical displacement:
d = (1/2) * g * t^2
Where:
- d is the vertical displacement (0.095 meters)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time the arrow is in the air
Rearranging the formula to solve for t, we have:
t^2 = (2 * d) / g
Substituting the given values, we get:
t^2 = (2 * 0.095) / 9.8
t^2 = 0.0194
Taking the square root of both sides, we find:
t ≈ 0.14 seconds
Therefore, the arrow was in the air for approximately 0.14 seconds.
b) To determine the initial velocity of the arrow when it left the bow, we can use the horizontal distance traveled by the arrow. Since the arrow is launched horizontally, the horizontal distance is equal to the horizontal component of the initial velocity (vx) multiplied by the time of flight (t).
d_horizontal = v_x * t
Where:
- d_horizontal is the horizontal distance (20.6 meters)
- v_x is the horizontal component of the initial velocity (the velocity when the bowstring is released)
- t is the time the arrow is in the air (0.14 seconds)
Rearranging the formula to solve for v_x, we have:
v_x = d_horizontal / t
Substituting the given values, we get:
v_x = 20.6 / 0.14
v_x ≈ 147.14 m/s
Therefore, the arrow was moving horizontally with an approximate speed of 147.14 m/s when it left the bow.
To solve this problem, we can use the equations of motion. Let's solve it step-by-step:
a) How long was the arrow in the air?
We will use the vertical motion equation:
y = ut + (1/2)gt^2
where:
y = vertical displacement
u = initial vertical velocity (zero in this case)
t = time of flight
g = acceleration due to gravity (-9.8 m/s^2)
Given that the arrow lands 0.095 meters below the target point, we can substitute these values into the equation:
-0.095 = 0 + (1/2)(-9.8)t^2
Simplifying the equation, we get:
-0.095 = -4.9t^2
Now let's solve for t:
t^2 = -0.095 / -4.9
t^2 = 0.01939
t = √0.01939
t ≈ 0.139 seconds
Therefore, the arrow was in the air for approximately 0.139 seconds.
b) How fast was the arrow moving when it left the bow?
To find the horizontal velocity of the arrow, we can use the horizontal motion equation:
x = vt
where:
x = horizontal displacement (20.6 meters in this case)
v = horizontal velocity
t = time of flight (0.139 seconds)
We can rearrange the equation to solve for v:
v = x / t
v = 20.6 / 0.139
v ≈ 148.20 m/s
Therefore, the arrow was moving at approximately 148.20 m/s when it left the bow.