An archer puts a 0.30kg arrow to the bowstring. An average force of 201N is exerted to draw the string back 0.8m. a. Assuming no frictional loss, with what speed does the arrow leave the bow? b. If the arrow is shot vertically into the air, how high will it rise?

a. The speed of the arrow as it leaves the bow can be calculated using the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.

The work done on the arrow is equal to the force applied to it times the distance over which the force is applied, or W = Fd.

Substituting the given values, we get W = 201N * 0.8m = 160.8 Joules.

This work is converted into kinetic energy, which is given by the formula KE = 1/2 mv^2, where m is the mass of the arrow and v is its speed.

Setting the work equal to the kinetic energy and solving for v, we get:

160.8J = 1/2 * 0.30kg * v^2

Solving for v, we get v = sqrt((2*160.8J)/0.30kg) = 34.64 m/s.

So, the arrow leaves the bow at a speed of 34.64 m/s.

b. The maximum height that the arrow will reach can be calculated using the conservation of energy. The kinetic energy of the arrow as it leaves the bow is converted into potential energy at the maximum height.

The potential energy is given by the formula PE = mgh, where m is the mass of the arrow, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.

Setting the kinetic energy equal to the potential energy and solving for h, we get:

160.8J = 0.30kg * 9.8 m/s^2 * h

Solving for h, we get h = 160.8J / (0.30kg * 9.8 m/s^2) = 54.67 m.

So, the arrow will rise to a maximum height of 54.67 m.