Asoft tennis ball is dropped onto a hard floor from a height of 2.50 m and rebounds back to a height of 2.15 m above the floor. gravity 10 m/s a). calculate its velocity just before it strikes the floor b). calculate its acceleration during contact with the floor if that contact lasts 2.50 ms c. how much time did the ball compress during its collision with the floor, assuming the floor is absolutely rigid? d). draw a detailed diagram of the situation 2+2+4+2 marks q6) suppose you first walk 18.5 m in a direction 28 west of north and then 25.7 m in a direction 44" south of west how far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? you must draw a vector diagram to represent the situation.

Ireally don’t know

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)

Mm: distances smaller than centimeters  km: for very large distanceshope this can i get brainliest

Your answer would be b. the total kinetic and potential energies of its particles