An archer shoots an arrow horizontally at a target 14 m away. The arrow is aimed directly at the center of the target, but it hits 59 cm lower. What was the initial speed of the arrow?

So, I am trying to find find v₀x. . .

Figured this one out already had to use the equation:

x = v₀x√(2h/g)

To find the initial speed of the arrow, we can use the horizontal motion formula. Since the arrow is shot horizontally, the initial vertical velocity of the arrow is 0.

The horizontal motion formula is:

d = v₀x * t

where:
d is the horizontal distance traveled (14 m),
v₀x is the initial horizontal velocity (which we are trying to find), and
t is the time of flight.

Since the vertical displacement is given (59 cm = 0.59 m), we can use the vertical motion formula to find the time of flight:

Δy = v₀y * t + (1/2) * g * t²

where:
Δy is the vertical displacement (-0.59 m),
v₀y is the initial vertical velocity (which is 0 since the arrow is shot horizontally), and
g is the acceleration due to gravity (approximately 9.8 m/s²).

Plugging in the values:

-0.59 = 0 * t + (1/2) * 9.8 * t²

Simplifying the equation:

-0.59 = (1/2) * 9.8 * t²

Multiplying both sides by 2:

-1.18 = 9.8 * t²

Dividing both sides by 9.8:

t² = -1.18 / 9.8

t² ≈ -0.120

Since time cannot be negative, we ignore the negative value.

t ≈ √(-0.120) (taking the square root of both sides)

t ≈ √0.120 (ignoring the negative square root)

t ≈ 0.346 seconds (approximated to three decimal places)

Now that we have the time of flight, we can use the horizontal motion formula to find the initial horizontal velocity:

14 = v₀x * 0.346

Dividing both sides by 0.346:

v₀x ≈ 40.5 m/s

Therefore, the initial speed (v₀x) of the arrow is approximately 40.5 m/s.