Well, let's see how this archery problem shoots itself in the foot with some physics humor!
To find the speed at which the arrow leaves the bow, we can use the principle of conservation of mechanical energy. The initial potential energy stored in the bowstring is converted into kinetic energy as the arrow is released.
First, we need to calculate the potential energy stored in the bowstring. We can do this using the formula:
Potential Energy (PE) = mgh,
where m is the mass of the arrow (0.26 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height difference (1.26 m) that the string is pulled back.
PE = (0.26 kg)(9.8 m/s^2)(1.26 m)
Now, let's convert this potential energy to kinetic energy using the equation:
Kinetic Energy (KE) = 0.5mv^2,
where v is the velocity of the arrow.
So, we set the potential energy equal to the kinetic energy:
PE = KE,
(mgh) = (0.5mv^2).
Now, we can solve for v:
v^2 = (2gh),
v = sqrt(2gh).
Substituting the values:
v = sqrt(2 * 9.8 m/s^2 * 1.26 m).
Calculating that:
v ≈ 5.01 m/s.
So, the arrow leaves the bow with a speed of approximately 5.01 m/s. That's one fast arrow flying through the air! Keep shooting for the stars, my friend!