An arrow is shot horizontally towards a target 20 m away. In traveling the first 5 m horizontally, the arrow falls 0.2 m. In traveling the next 5 m horizontally, what will the arrow additionally fall?
To determine how much the arrow will additionally fall in the next 5 meters, we need to first calculate the vertical displacement of the arrow during the first 5 meters. We know that the arrow falls 0.2 meters during this distance.
Next, we can use the concept of uniform acceleration due to gravity to calculate the vertical displacement of the arrow in the next 5 meters. Since the arrow is shot horizontally, there is no horizontal acceleration, and only vertical acceleration due to gravity is acting on it.
The equation we can use to calculate the vertical displacement is given by:
y = ut + (1/2)gt^2
Where:
y represents the vertical displacement
u represents the initial vertical velocity (which is 0 since the arrow is shot horizontally)
g represents the acceleration due to gravity (9.8 m/s^2)
t represents the time
Since the arrow travels the first 5 meters horizontally, which takes the same amount of time to travel the next 5 meters horizontally, we can use this equation to calculate the vertical displacement during the next 5 meters by substituting the distance of 5 meters for y:
5 = 0 + (1/2)(9.8)t^2
To solve for t^2, we can rearrange the equation:
9.8t^2 = 10
t^2 = 10 / 9.8
t^2 ≈ 1.02
Now, we can substitute this value for t^2 back into the equation to find the vertical displacement during the next 5 meters:
y = 0 + (1/2)(9.8)(1.02)
y ≈ 4.99
Therefore, the arrow will additionally fall approximately 4.99 meters in the next 5 meters horizontally.