An arrow fired horizontally at 45 m/s travels 26 m horizontally. From what height was it fired?
26=45t
t=26/45 seconds
height=1/2 g t^2
put in t, solve for height
To determine the height from which the arrow was fired, we can use the kinematic equations of motion.
Firstly, we need to identify the known values:
- Initial horizontal velocity (v_x) = 45 m/s (the arrow is fired horizontally, so its initial vertical velocity is zero)
- Horizontal displacement (Δx) = 26 m
- Acceleration due to gravity (g) = 9.8 m/s²
We are trying to find the initial vertical displacement (Δy) or the height from which the arrow was fired.
We can use the equation for horizontal displacement:
Δx = v_x * t (equation 1)
Since there is no vertical acceleration, the time taken for the arrow to reach the ground horizontally (t) is the same as the time taken for it to fall vertically to the ground. Therefore, we can use the equation for vertical displacement:
Δy = (1/2) * g * t² (equation 2)
Now, we can solve for the time taken (t) using equation 1:
t = Δx / v_x
t = 26 m / 45 m/s ≈ 0.5778 s
Substituting this time value into equation 2, we can solve for the vertical displacement (Δy):
Δy = (1/2) * g * (0.5778 s)²
Δy ≈ (1/2) * 9.8 m/s² * 0.3348 s²
Δy ≈ 1.6349 m
Therefore, the arrow was fired from a height of approximately 1.6349 meters.