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JEE Advanced 2019 Paper 2
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© examsnet.com
Question : 10
Total: 54
A ball is thrown from ground at angle θ with horizontal and with an initial speed
u
0
For the resulting projectile motion, the magnitude of average velocity of the ball up to the point when it hits the ground for the first time is
V
1
. After hitting the ground, the ball rebounds at the same angle θ but with a reduced speed of
u
0
α
.Its motion continues for a long time as shown in figure. If the magnitude of average velocity of the ball for entire duration of motion is 0.8
V
1
the value of α is______
Your Answer:
Validate
Solution:
R
1
=
u
0
2
sin
2
θ
g
R
2
=
1
α
2
(
u
0
2
sin
2
θ
g
)
=
R
1
α
2
t
1
=
2
u
0
sin
θ
g
t
2
=
t
1
α
.
.
.
.
.
.
.
.
T
=
t
1
+
t
2
+
.
.
.
.
=
t
1
(
1
1
−
1
α
)
=
α
t
1
α
−
1
Average speed
=
R
1
t
1
Average speed
=
R
T
α
=
4
© examsnet.com
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