AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper

The time and speed of the ball to reach on the ground is,
t0=√

2h

g

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=

v1

g

The time taken between first and second collision is 2t1 .
Similarly the time taken between second and third collision is, 2t3=

2v2

g

Thus the total time before it ceaser to rebound is,
T=t0+2t1+2t2+...
T=t0+

2v1

g

+

2v2

g

+....
T=t0+

2ev0

g

+

2e2v0

g

+... ( As, v2=ev1=e2v0)
T=√

2h

g

[1+2e(1+e+e2+e3+....)] ( As, v0=√

2h

g

)
Solve further,
T=√

2h

g

[1+2e(

1

1−e

)]
T=√

2h

g

[

1+e

1−e

]
T=√

2(180)

10

(

1+0.5

1−0.5

)
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(

1+e2

1−e2

)
Substitute values,
H=180(

1+ 1 4

1− 1 4

)
H=300m
The average speed is,
|v|=

H

T

|v|=

300

18

|v|=

50

3

m∕s
The average velocity is,
v=

180

18

v=10m∕s