You have a friend who is an archer and she wants to know how fast the arrows leave her bow. You have her launch an arrow horizontally, 1.5 m above the ground, and find that the arrow lands 33 m away. What is the initial velocity of the arrow as it leaves the bow?

To determine the initial velocity of the arrow, we can use the principles of projectile motion. When an arrow is launched horizontally, it only has an initial horizontal velocity and no initial vertical velocity.

Given:
Initial vertical position (y): 1.5 m
Horizontal distance traveled (x): 33 m

Now, we can make use of the equation that relates the horizontal distance traveled by an object in projectile motion to its initial horizontal velocity:

x = v₀x * t

where:
x is the horizontal distance traveled (33 m),
v₀x is the initial horizontal velocity (the value we want to find),
t is the time of flight of the arrow.

Since the arrow is launched horizontally, the time of flight is the same as the time it takes for the arrow to fall vertically from its initial height of 1.5 m to the ground.

We can determine the time of flight using the equation that relates the vertical displacement of an object in free fall to the gravitational acceleration:

y = (1/2) * g * t²

where:
y is the vertical displacement (1.5 m),
g is the acceleration due to gravity (approximately 9.8 m/s²),
t is the time of flight (the value we want to find).

Rearranging the equation and solving for t:

t = √(2 * y / g)

Given the initial height (y) is 1.5 m and gravitational acceleration (g) is approximately 9.8 m/s², we can calculate the value of t.

t = √(2 * 1.5 / 9.8)
t ≈ √0.306 ≈ 0.553 s

Now, we can substitute the known values of x and t into the first equation to find the initial horizontal velocity (v₀x):

33 m = v₀x * 0.553 s

Solving for v₀x:

v₀x ≈ 33 m / 0.553 s ≈ 59.67 m/s

Therefore, the initial velocity of the arrow as it leaves the bow is approximately 59.67 m/s.