You drop a cherry pit out the car window 1.0 m above the ground while traveling down the road at a constant speed of 18 m/s. how far, horizontally, from the initial dropping point will the pit hit the ground? if the car continues to travel at the same speed, where will the car be in relation to the pit when it lands?

No because scientist are still changing there mines today so they are not correct to this day

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)