Starting with an initial speed of 6.95 m/s at a height of 0.150 m, a 2.84-kg ball swings downward and strikes a 4.97-kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the 2.84-kg ball just before impact. (b) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 2.84-kg ball just after the collision. (c) Assuming that the collision is elastic, find the velocity (magnitude and direction) of the 4.97-kg ball just after the collision. (d) How high does the 2.84-kg ball swing after the collision, ignoring air resistance

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The figure is missing but we can answer anyway.the correct answer is "releasing the string". in fact, when the person is holding the bow and the arrow, the kinetic energy of the arrow is still zero (because the arrow is not moving). the moment the arrow is released, it acquires a velocity v, so its kinetic energy will become different from zero: where m is the mass of the arrow and v its velocity.  all the other options, instead, do not increase the kinetic energy of the system: -  pulling farther back on the string : this will just increase the elastic potential energy of the system-  holding the arrow higher: this will increase the gravitational potential energy of the system (which is given by u=mgh, where h is the height above the ground)-  decreasing the mass of the arrow: this will decrease the gravitational potential energy of the system (which is given by u=mgh, where m is the mass)