An archer shoots an arrow horizontally at a target 12 m away. The arrow is aimed directly at the center of the target, but it hits 56 cm lower. What was the initial speed of the arrow? (Neglect air resistance.)
8.29
To find the initial speed of the arrow, we can use the equations of motion for projectile motion. We know that the arrow is shot horizontally, so its initial vertical velocity is zero.
Let's break down the given information:
Target distance (horizontal displacement, x): 12 m
Vertical displacement, y: -56 cm = -0.56 m (it hits 56 cm lower)
In projectile motion, the horizontal and vertical motions are independent of each other. The equations we'll be using are:
Horizontal motion: x = v0 * t
Vertical motion: y = v0y * t + (1/2) * g * t^2
Since the horizontal velocity (v0x) is constant, and in this case, it's zero, we don't need to consider it in the calculation.
We need to find the initial vertical velocity (v0y) of the arrow, which is the speed at which the arrow is launched.
Let's start with the vertical motion equation:
y = v0y * t + (1/2) * g * t^2
Since the initial vertical velocity is zero, the equation simplifies to:
y = (1/2) * g * t^2
We can isolate t^2:
t^2 = (2 * y) / g
Next, we'll use the horizontal motion equation:
x = v0 * t
Since the initial horizontal velocity is zero, the equation simplifies to:
x = 0
We can see that x is zero because the arrow was shot horizontally and we're measuring the distance relative to the starting point of the arrow.
Now, let's solve for time (t):
t = x / v0
Since x is zero, we find that t is also zero.
Now, we have the time (t) and can substitute it back into the vertical motion equation:
y = (1/2) * g * t^2
y = (1/2) * g * 0^2
y = 0
We can see that the vertical displacement (y) is zero, but the given information states that the arrow hits 56 cm lower (negative displacement).
So, there must be an error in the problem statement or some missing information. Please double-check the information provided or provide additional details.