Answer:
The arrow will fall an additional 0.6 m.
Step-by-step explanation:
The height of the arrow can be calculated using the following kinematic equation of a falling object:
y = y0 + v0y · t + 1/2 · g · t²
Where:
y = height of the arrow at time "t"
y0 = initial height.
v0y = initial vertical velocity.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
t = time.
Since the arrow is shot horizontally, initially, it does not have a vertical velocity, so, v0y = 0.
If we place the origin of the frame of reference at the throwing point, y0 is also 0. Then:
y = 1/2 · g · t²
First, let´s calculate how much time it takes for the arrow to travel 5 m. It will be the time it takes the arrow to fall 0.2 m. Then, using the equation of height:
y = 1/2 · g · t²
-0.2 m = 1/2 · (-9.8 m/s²) · t² (notice that we consider the upward direction as positive and the origin of the frame of reference is at the throwing point).
-0.2 m / -(1/2 · 9.8 m/s²) = t²
t = 0.2 s
If we neglect air resistance, the horizontal velocity of the arrow will be constant because there is no force acting in the horizontal direction on the arrow. Then, the arrow will travel the next 5 m in another 0.2 s. Then, let´s find the height of the arrow at t = 0.4 s.
y = 1/2 · g · t²
Instead of t = 0.4 s I will use t = 2 · √(-0.2 m / -(1/2 · 9.8 m/s²)) (this expression was obtained above) to avoid error because of rounding:
y = 1/2 · (-9.8² m/s) · (2 · √(-0.2 m / -(1/2 · 9.8 m/s²)))²
y = -0.8 m
The height of the arrow when it travels another 5 m will be -0.8 m (0.8 m below the throwing point). It means that the arrow will fall an additional 0.6 m.
