Speed after #n# rebounds ?

1 Answer
Feb 27, 2017

Answer:

#h_n =(v_0^2 epsilon^(2n))/(2g)#

Explanation:

After #n# rebounds the ascending speed is

#v_n = v_0 epsilon ^n# where #v_0 = sqrt((g h)/2)# is the initial speed.

with that initial speed, using the mechanical energy conservation relationship

#1/2m v_n^2=m g h_n# or #h_n = v_n^2/(2g) = (v_0^2 epsilon^(2n))/(2g)#

then the equation for #h_n# is

#h_n =(v_0^2 epsilon^(2n))/(2g)#