An archer shoots and arrow horizontally at a target 15m away. The arrow is aimed directly at the center of the target, but it hits 52 cm lower.how long did it take for the arrow to reach the target

How long does it take to fall 0.52 meters?

.52 = (1/2) g t^2

1.04 = 9.81 t^2

t = 0.326 seconds

To determine the time it took for the arrow to reach the target, we can use the horizontal distance and the vertical displacement of the arrow.

Given:
Horizontal distance (d) = 15 m
Vertical displacement (h) = 52 cm = 0.52 m

We can use the kinematic equation to solve for time (t):

h = 1/2 * g * t^2

By substituting the known values:

0.52 = 1/2 * 9.8 * t^2

Simplifying the equation:

0.52 = 4.9 * t^2

Dividing both sides by 4.9:

0.52/4.9 = t^2

0.10612 = t^2

Taking the square root of both sides:

t ≈ 0.326 s

Therefore, it took approximately 0.326 seconds for the arrow to reach the target.

To determine how long it took for the arrow to reach the target, we can use the equations of motion. Since the arrow is shot horizontally, its initial vertical velocity will be zero. We need to find the time when the arrow reaches the target, knowing that it falls 52 cm (or 0.52 m) lower than the aim point.

Using the equation for vertical displacement:

Δy = v₀y * t + (1/2) * a * t²

Where:
Δy = vertical displacement (in this case, -0.52 m)
v₀y = initial vertical velocity (which is zero since the arrow is shot horizontally)
a = acceleration due to gravity (approximately -9.8 m/s²)
t = time

Plugging in the given values, the equation becomes:

-0.52 = 0 * t + (1/2) * (-9.8) * t²

Simplifying the equation further:

-0.52 = (-4.9) * t²

Dividing both sides by -4.9:

t² = 0.52 / 4.9

t² ≈ 0.106

Taking the square root of both sides:

t ≈ √0.106

t ≈ 0.326 s

Therefore, it took approximately 0.326 seconds for the arrow to reach the target.

-7.67