An archer shoots an arrow horizontally at a target 14 m away. The arrow is aimed directly at the center of the target, but it hits 52 cm lower.
1-What was the initial speed of the arrow?
find the time it takes an object to fall 52cm.
then, velocity=14m/time
46 m/s
To find the initial speed of the arrow, we can use the horizontal distance and the vertical displacement.
Given:
Horizontal distance (x) = 14 m
Vertical displacement (y) = -52 cm = -0.52 m
We can use the kinematic equation for vertical motion:
y = (1/2)gt^2
where:
y = vertical displacement
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time of flight
First, let's find the time of flight:
y = (1/2)gt^2
-0.52 = (1/2)(9.8)t^2
-1.04 = (9.8)t^2
t^2 = -1.04/9.8
t^2 = -0.10612
t ≈ √(-0.10612)
t ≈ 0.326 seconds
Now that we have the time of flight, we can find the initial vertical velocity (vy) using the kinematic equation:
y = vyt - (1/2)gt^2
-0.52 = vy(0.326) - (1/2)(9.8)(0.326)^2
-0.52 = 0.326vy - 0.522014
0.326vy = -0.52 + 0.522014
0.326vy = 0.002014
vy ≈ 0.00617 m/s
Since the arrow was shot horizontally, the vertical velocity component is zero. Therefore, the initial speed of the arrow is equal to the horizontal velocity (Vx).
Using the formula for constant horizontal velocity:
Vx = x/t
Vx = 14 / 0.326
Vx ≈ 42.94 m/s
Therefore, the initial speed of the arrow is approximately 42.94 m/s.
To find the initial speed of the arrow, we can use the concept of projectile motion. Since the arrow is shot horizontally, the initial vertical velocity will be zero. The arrow's motion can be divided into the vertical and horizontal components.
The horizontal distance traveled by the arrow is 14 m, and the vertical distance it falls below the target is 52 cm, which is 0.52 m.
In the vertical direction, the only force acting on the arrow is gravity. We can use the equation of motion for vertical motion:
y = ut + (1/2)gt^2
Where:
y = vertical displacement (0.52 m)
u = initial vertical velocity (0 m/s)
t = time of flight
g = acceleration due to gravity (-9.8 m/s^2)
Since the vertical displacement is negative (falling below the target), the equation becomes:
0.52 = (1/2)(-9.8)t^2
Simplifying the equation, we get:
0.52 = -4.9t^2
Rearranging the equation, we have:
t^2 = 0.52 / -4.9
t^2 = -0.1061
Taking the square root of both sides, we find:
t ≈ 0.3262 s
Now, we can use the horizontal distance and the time of flight to find the initial speed of the arrow. The horizontal distance is 14 m, and the time of flight is approximately 0.3262 seconds.
The equation of motion for horizontal motion is:
x = ut
Where:
x = horizontal displacement (14 m)
u = initial horizontal velocity (unknown)
t = time of flight (0.3262 s)
Rearranging the equation, we get:
u = x / t
Substituting the values, we have:
u = 14 m / 0.3262 s
Calculating the result, we find:
u ≈ 42.92 m/s
Therefore, the initial speed of the arrow is approximately 42.92 m/s.