a ball is dropped on to the floor from a height of 10 m. it rebounds to aheight of 2.5m. if thw ball is in contact with the floor for 0.01seconds. find averge accelaration

consider upwards as positive and down wards as negative. it falls from a height of 10 m. so on reaching the ground it has a velocity of squareroot(2*g*10) = 10*1.414 = 14.14m/s in the downward direction. (taking g=10m/s) as it rebounds to a height of 2.5m its velocity soon after leaving the ground is squareroot(2*g*2.5) = 5*1.414 = 7.07 m/s in the upward direction both the above values can be found from the equation v^2 - u^2 = 2*a*s we have average accleration = change in velocity / time taken = 7.07 - (-14.14) / 0.01 = 2121 m/s^2 that is 2121 m/s^2 in the upward direction.