The time and speed of the ball to reach on the ground is,
t0=√ and
v0=√2gh After the first collision, the speed of the ball is,
v1=ev0=e√2gh The time when ball stops the motion is,
v=u+at 0=v1−gt1 t1= The time taken between first and second collision is
2t1 .
Similarly the time taken between second and third collision is,
2t3= Thus the total time before it ceaser to rebound is,
T=t0+2t1+2t2+... T=t0+++.... T=t0+++... ( As,
v2=ev1=e2v0) T=√[1+2e(1+e+e2+e3+....)] ( As,
v0=√) Solve further,
T=√[1+2e()] T=√[] T=√() T=18s Similarly the total distance in the entire motion of ball is,
H=h0+2h1+2h2+... H=h+2e2h+2e2h+...... H=h[1+2e2(1+e2+e4+......)] H=h() Substitute values,
H=180() H=300m The average speed is,
|v|= |v|= |v|=m∕s The average velocity is,
v= v=10m∕s