An archer shoots an arrow horizontally at a target 15 m away. the arrow is aimed directly at the centre of the target, but it hits 15 cm lower. what would be its initial speed.
how long does it take to fall 15cm? 4.9t^2 = 0.15
Use that time t, and note that the horizontal speed is 15/t m/s
To find the initial speed of the arrow, we can use the kinematic equation for vertical motion. The equation we need is:
y = Vi*t + (1/2)*a*t^2
Where:
- y is the vertical displacement (15 cm = 0.15 m in this case)
- Vi is the initial vertical velocity (what we want to find)
- t is the time it takes for the arrow to reach the target (which is the same time it takes for the arrow to travel horizontally)
- a is the acceleration due to gravity (-9.8 m/s^2 for downward motion)
Since the arrow is shot horizontally, the initial vertical velocity (Vi) is 0 m/s. Therefore, the equation simplifies to:
y = (1/2)*a*t^2
Now let's rearrange the equation to solve for t:
t^2 = (2*y) / a
t^2 = (2 * 0.15) / 9.8
t^2 ≈ 0.0306
t ≈ √0.0306
t ≈ 0.175 s (rounded to three decimal places)
Now that we know the time it takes for the arrow to reach the target horizontally, we can use this time in the horizontal motion equation:
x = Vx*t
We are given that the horizontal distance traveled (x) is 15 m. Since the arrow is shot horizontally, the initial horizontal velocity (Vx) is what we want to find. Rearranging the equation:
Vx = x / t
Vx = 15 / 0.175
Vx ≈ 85.714 m/s (rounded to three decimal places)
Therefore, the initial speed of the arrow is approximately 85.714 m/s.