A ball of moist clay falls 15.0m to the ground. It is in contact with the ground for 20ms before stopping?

To find the velocity when it reached the ground, use

#v^2 = u^2 + 2*a*s#

#v^2 = 0^2 + 2*9.8 m/s^2*15.0 m#

#v^2 = 294 m^2/s^2#

#v = 17.15 m/s#

(a) To find the average acceleration over the time of 0.020 s, use

#v = u + a*t#

Since #17.15 m/s# was the initial velocity and it was downward, let the #u = -17.15 m/s#.

#0 = -17.15 m/s + a*0.020 s#

#a_"ave" = 17.15 m/s /0.020 s = 857 m/s^2#

(b) Since #a_"ave"# turned out to be positive and the falling velocity was negative, the acceleration must have been upwards. And that makes sense, since to stop something, you need a force in the direction opposite the existing velocity. And from #vecF=m*veca#, we can see that acceleration is in the same direction as the force acting on the object.

I hope this helps,
Steve

The velocity of the clay ball as it hits the ground is found from the kinematic expression

#v^2 - u^2 = 2gh#

Substituting given values we get

#v^2 - 0^2 = 2xx9.81xx15.0#
#=>v=17.16\ ms^-1#

After it hits ground it is brought to stop. Acceleration can be found from the kinematic expression

#v=u+at#

Substituting various values we get

#0=17.16+a_"ave"xx(20xx10^-3)#
#=>a_"ave"=-17.16/(20xx10^-3)#
#=>a_"ave"=-857.8\ ms^-2#

(a) #857.8\ ms^-2#
(b) #-ve# sign shows us that it is retardation and opposes the motion. The motion of the ball is in the downward direction. Hence, average acceleration acts in up direction.