JEE Main Previous Year Questions (2025): Motion in a Plane | Physics for JEE Main & Advanced PDF Download

 Page 1


JEE Main Previous Year Qs (2025): 
Motion in a Plane 
Q1: A particle is projected at an angle of ????
°
 from horizontal at a speed 
of ???? ?? /?? . The height traversed by the particle in the first second is ?? ?? 
and height traversed in the last second, before it reaches the 
maximum height, is ?? ?? . The ratio ?? ?? :?? ?? is ____ . 
[Take, ?? = ???? ?? /?? ?? ] 
JEE Main 2025 (Online) 22nd January Morning Shift 
Answer: 5 
Solution: 
 
 
?? ?? = ?? ?? - ???? 
?? = 30- 10?? ? ?? = 3?? 
We know, ?? ?? = ?? +
?? 2
(2?? - 1) 
So, h
0
= ?? 1
= 30-
10
2
(2 × 1 - 1) = 30- 5 = 25 m 
and h
1
= ?? ?? 3
= 30-
10
2
(2 × 3 - 1) = 30- 25 = 5 
Hence, 
h
0
h
1
=
25
5
= 5 
 
Q2: The maximum speed of a boat in still water is ???? ???? /?? . Now this 
boat is moving downstream in a river flowing at ?? ???? /?? . A man in the 
boat throws a ball vertically upwards with speed of ???? ?? /?? . Range of 
the ball as observed by an observer at rest on the bank is ____ cm. (Take 
?? = ???? ?? /?? ?? ) 
JEE Main 2025 (Online) 29th January Morning Shift 
Page 2


JEE Main Previous Year Qs (2025): 
Motion in a Plane 
Q1: A particle is projected at an angle of ????
°
 from horizontal at a speed 
of ???? ?? /?? . The height traversed by the particle in the first second is ?? ?? 
and height traversed in the last second, before it reaches the 
maximum height, is ?? ?? . The ratio ?? ?? :?? ?? is ____ . 
[Take, ?? = ???? ?? /?? ?? ] 
JEE Main 2025 (Online) 22nd January Morning Shift 
Answer: 5 
Solution: 
 
 
?? ?? = ?? ?? - ???? 
?? = 30- 10?? ? ?? = 3?? 
We know, ?? ?? = ?? +
?? 2
(2?? - 1) 
So, h
0
= ?? 1
= 30-
10
2
(2 × 1 - 1) = 30- 5 = 25 m 
and h
1
= ?? ?? 3
= 30-
10
2
(2 × 3 - 1) = 30- 25 = 5 
Hence, 
h
0
h
1
=
25
5
= 5 
 
Q2: The maximum speed of a boat in still water is ???? ???? /?? . Now this 
boat is moving downstream in a river flowing at ?? ???? /?? . A man in the 
boat throws a ball vertically upwards with speed of ???? ?? /?? . Range of 
the ball as observed by an observer at rest on the bank is ____ cm. (Take 
?? = ???? ?? /?? ?? ) 
JEE Main 2025 (Online) 29th January Morning Shift 
Answer: 2000 
Solution: 
The first step is to determine the horizontal velocity of the ball as seen by an observer on the bank. 
The velocity of the boat in still water is 27 km/h. Since the boat is moving downstream and the 
river flows at 9 km/h, the actual speed of the boat relative to the bank becomes: 
?? boat 
= 27+ 9 = 36 km/h 
Convert this speed from km/h to m/s by using the conversion factor 1 km/h =
5
18
 m/s : 
?? boat 
= 36 ×
5
18
= 10 m/s 
This is also the horizontal velocity of the ball as seen by the observer on the bank. 
Next, calculate the time the ball spends in the air. The initial vertical velocity of the ball is 10 m/s. 
The time to reach the maximum height ?? up 
 is given by: 
?? up 
=
?? initial 
?? =
10
10
= 1 s 
The total time of flight ?? total 
 is twice the time to reach the maximum height: 
?? total 
= 2× 1 = 2 s 
The horizontal range ?? of the ball is the horizontal velocity multiplied by the total time of flight: 
?? = ?? horizontal 
 × ?? total 
= 10× 2 = 20 m 
Finally, convert this range into centimeters: 
?? = 20× 100= 2000 cm 
Thus, the range of the ball as observed by an observer at rest on the bank is 2000 cm . 
 
Q3: A ball of mass ???????? is projected with velocity ???? ?? /?? at ????
°
 with 
horizontal. The decrease in kinetic energy of the ball during the 
motion from point of projection to highest point is 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 20 J 
B. 5 J 
C. 15 J 
D. zero 
Answer: C 
Solution: 
 
Page 3


JEE Main Previous Year Qs (2025): 
Motion in a Plane 
Q1: A particle is projected at an angle of ????
°
 from horizontal at a speed 
of ???? ?? /?? . The height traversed by the particle in the first second is ?? ?? 
and height traversed in the last second, before it reaches the 
maximum height, is ?? ?? . The ratio ?? ?? :?? ?? is ____ . 
[Take, ?? = ???? ?? /?? ?? ] 
JEE Main 2025 (Online) 22nd January Morning Shift 
Answer: 5 
Solution: 
 
 
?? ?? = ?? ?? - ???? 
?? = 30- 10?? ? ?? = 3?? 
We know, ?? ?? = ?? +
?? 2
(2?? - 1) 
So, h
0
= ?? 1
= 30-
10
2
(2 × 1 - 1) = 30- 5 = 25 m 
and h
1
= ?? ?? 3
= 30-
10
2
(2 × 3 - 1) = 30- 25 = 5 
Hence, 
h
0
h
1
=
25
5
= 5 
 
Q2: The maximum speed of a boat in still water is ???? ???? /?? . Now this 
boat is moving downstream in a river flowing at ?? ???? /?? . A man in the 
boat throws a ball vertically upwards with speed of ???? ?? /?? . Range of 
the ball as observed by an observer at rest on the bank is ____ cm. (Take 
?? = ???? ?? /?? ?? ) 
JEE Main 2025 (Online) 29th January Morning Shift 
Answer: 2000 
Solution: 
The first step is to determine the horizontal velocity of the ball as seen by an observer on the bank. 
The velocity of the boat in still water is 27 km/h. Since the boat is moving downstream and the 
river flows at 9 km/h, the actual speed of the boat relative to the bank becomes: 
?? boat 
= 27+ 9 = 36 km/h 
Convert this speed from km/h to m/s by using the conversion factor 1 km/h =
5
18
 m/s : 
?? boat 
= 36 ×
5
18
= 10 m/s 
This is also the horizontal velocity of the ball as seen by the observer on the bank. 
Next, calculate the time the ball spends in the air. The initial vertical velocity of the ball is 10 m/s. 
The time to reach the maximum height ?? up 
 is given by: 
?? up 
=
?? initial 
?? =
10
10
= 1 s 
The total time of flight ?? total 
 is twice the time to reach the maximum height: 
?? total 
= 2× 1 = 2 s 
The horizontal range ?? of the ball is the horizontal velocity multiplied by the total time of flight: 
?? = ?? horizontal 
 × ?? total 
= 10× 2 = 20 m 
Finally, convert this range into centimeters: 
?? = 20× 100= 2000 cm 
Thus, the range of the ball as observed by an observer at rest on the bank is 2000 cm . 
 
Q3: A ball of mass ???????? is projected with velocity ???? ?? /?? at ????
°
 with 
horizontal. The decrease in kinetic energy of the ball during the 
motion from point of projection to highest point is 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 20 J 
B. 5 J 
C. 15 J 
D. zero 
Answer: C 
Solution: 
 
 
 
k
i
=
1
2
mv
2
 
k
f
=
1
2
 m(vcos 60
°
)
2
=
1
8
?? v
2
 
?k = k
i
- k
f
=
3
8
mv
2
=
3
8
× 0.1 × 400= 15 J 
 
Q4: An object of mass ' ?? ' is projected from origin in a vertical xy 
plane at an angle ????
°
 with the x - axis with an initial velocity ?? ?? . The 
magnitude and direction of the angular momentum of the object with 
respect to origin, when it reaches at the maximum height, will be [ ?? is 
acceleration due to gravity] 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 
?? ?? ?? 
3
2v2?? along negative ?? -axis 
B. 
?? ?? ?? 3
2v2?? along positive ?? -axis 
C. 
?? ?? ?? 3
4v2?? along positive ?? -axis 
D. 
?? ?? ?? 3
4v2?? along negative z -axis 
Answer: D 
Solution: 
?? 0
???? =
?? 0
v2
??ˆ +
?? 0
v2
??ˆ. 
At maximum height, the vertical component of the velocity becomes zero. Setting the vertical 
Page 4


JEE Main Previous Year Qs (2025): 
Motion in a Plane 
Q1: A particle is projected at an angle of ????
°
 from horizontal at a speed 
of ???? ?? /?? . The height traversed by the particle in the first second is ?? ?? 
and height traversed in the last second, before it reaches the 
maximum height, is ?? ?? . The ratio ?? ?? :?? ?? is ____ . 
[Take, ?? = ???? ?? /?? ?? ] 
JEE Main 2025 (Online) 22nd January Morning Shift 
Answer: 5 
Solution: 
 
 
?? ?? = ?? ?? - ???? 
?? = 30- 10?? ? ?? = 3?? 
We know, ?? ?? = ?? +
?? 2
(2?? - 1) 
So, h
0
= ?? 1
= 30-
10
2
(2 × 1 - 1) = 30- 5 = 25 m 
and h
1
= ?? ?? 3
= 30-
10
2
(2 × 3 - 1) = 30- 25 = 5 
Hence, 
h
0
h
1
=
25
5
= 5 
 
Q2: The maximum speed of a boat in still water is ???? ???? /?? . Now this 
boat is moving downstream in a river flowing at ?? ???? /?? . A man in the 
boat throws a ball vertically upwards with speed of ???? ?? /?? . Range of 
the ball as observed by an observer at rest on the bank is ____ cm. (Take 
?? = ???? ?? /?? ?? ) 
JEE Main 2025 (Online) 29th January Morning Shift 
Answer: 2000 
Solution: 
The first step is to determine the horizontal velocity of the ball as seen by an observer on the bank. 
The velocity of the boat in still water is 27 km/h. Since the boat is moving downstream and the 
river flows at 9 km/h, the actual speed of the boat relative to the bank becomes: 
?? boat 
= 27+ 9 = 36 km/h 
Convert this speed from km/h to m/s by using the conversion factor 1 km/h =
5
18
 m/s : 
?? boat 
= 36 ×
5
18
= 10 m/s 
This is also the horizontal velocity of the ball as seen by the observer on the bank. 
Next, calculate the time the ball spends in the air. The initial vertical velocity of the ball is 10 m/s. 
The time to reach the maximum height ?? up 
 is given by: 
?? up 
=
?? initial 
?? =
10
10
= 1 s 
The total time of flight ?? total 
 is twice the time to reach the maximum height: 
?? total 
= 2× 1 = 2 s 
The horizontal range ?? of the ball is the horizontal velocity multiplied by the total time of flight: 
?? = ?? horizontal 
 × ?? total 
= 10× 2 = 20 m 
Finally, convert this range into centimeters: 
?? = 20× 100= 2000 cm 
Thus, the range of the ball as observed by an observer at rest on the bank is 2000 cm . 
 
Q3: A ball of mass ???????? is projected with velocity ???? ?? /?? at ????
°
 with 
horizontal. The decrease in kinetic energy of the ball during the 
motion from point of projection to highest point is 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 20 J 
B. 5 J 
C. 15 J 
D. zero 
Answer: C 
Solution: 
 
 
 
k
i
=
1
2
mv
2
 
k
f
=
1
2
 m(vcos 60
°
)
2
=
1
8
?? v
2
 
?k = k
i
- k
f
=
3
8
mv
2
=
3
8
× 0.1 × 400= 15 J 
 
Q4: An object of mass ' ?? ' is projected from origin in a vertical xy 
plane at an angle ????
°
 with the x - axis with an initial velocity ?? ?? . The 
magnitude and direction of the angular momentum of the object with 
respect to origin, when it reaches at the maximum height, will be [ ?? is 
acceleration due to gravity] 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 
?? ?? ?? 
3
2v2?? along negative ?? -axis 
B. 
?? ?? ?? 3
2v2?? along positive ?? -axis 
C. 
?? ?? ?? 3
4v2?? along positive ?? -axis 
D. 
?? ?? ?? 3
4v2?? along negative z -axis 
Answer: D 
Solution: 
?? 0
???? =
?? 0
v2
??ˆ +
?? 0
v2
??ˆ. 
At maximum height, the vertical component of the velocity becomes zero. Setting the vertical 
velocity to zero: 
?? 0
v2
- ???? = 0 ? ?? =
?? 0
?? v2
. 
The coordinates at this time are: 
?? =
?? 0
v2
?? =
?? 0
v2
·
?? 0
?? v2
=
?? 0
2
2?? , 
?? =
?? 0
v2
?? -
1
2
?? ?? 2
=
?? 0
2
2?? -
1
2
?? ?? 0
2
2?? 2
=
?? 0
2
2?? -
?? 0
2
4?? =
?? 0
2
4?? . 
Thus, the position vector at maximum height is: 
?? =
?? 0
2
2?? ??ˆ +
?? 0
2
4?? ??ˆ. 
At maximum height, the only nonzero component of velocity is the horizontal one: 
?? =
?? 0
v2
??ˆ. 
The angular momentum about the origin is given by: 
?? ? 
= ?? ?? × ?? . 
Writing out the cross product: 
?? × ?? = (
?? 0
2
2?? ??ˆ +
?? 0
2
4?? ??ˆ) ×
?? 0
v2
??ˆ. 
Since the cross product of ??ˆ with itself is zero and: 
??ˆ × ??ˆ = -?? ˆ
, 
we have: 
?? × ?? =
?? 0
2
4?? ·
?? 0
v2
(-?? ˆ
) = -
?? 0
3
4v2?? ?? ˆ
. 
Thus, the angular momentum is: 
?? ? 
= -
?? ?? 0
3
4v2?? ?? ˆ
. 
This means the magnitude is: 
?? ?? 0
3
4v2?? , 
and its direction is along the negative ?? ˆ
 (or negative ?? -axis). 
 
Q5: The position vector of a moving body at any instant of time is 
given as ?? = (?? ?? ?? ??ˆ - ???? ??ˆ)?? . The magnitude and direction of velocity at 
?? = ?? ?? is, 
JEE Main 2025 (Online) 24th January Evening Shift 
Options: 
A. 5v17 m/s, making an angle of tan
-1
 4 with - ve Y axis 
B. 5v15 m/s, making an angle of tan
-1
 4 with + ve ?? axis 
C. 5v17 m/s, making an angle of tan
-1
 4 with + ve ?? axis 
D. 5v15 m/s, making an angle of tan
-1
 4 with - ve ?? axis 
Answer: A 
Solution: 
r = 5t
2
i ˆ - 5t
ˆ
 
v ? = 10ti ˆ - 5j ˆ 
Page 5


JEE Main Previous Year Qs (2025): 
Motion in a Plane 
Q1: A particle is projected at an angle of ????
°
 from horizontal at a speed 
of ???? ?? /?? . The height traversed by the particle in the first second is ?? ?? 
and height traversed in the last second, before it reaches the 
maximum height, is ?? ?? . The ratio ?? ?? :?? ?? is ____ . 
[Take, ?? = ???? ?? /?? ?? ] 
JEE Main 2025 (Online) 22nd January Morning Shift 
Answer: 5 
Solution: 
 
 
?? ?? = ?? ?? - ???? 
?? = 30- 10?? ? ?? = 3?? 
We know, ?? ?? = ?? +
?? 2
(2?? - 1) 
So, h
0
= ?? 1
= 30-
10
2
(2 × 1 - 1) = 30- 5 = 25 m 
and h
1
= ?? ?? 3
= 30-
10
2
(2 × 3 - 1) = 30- 25 = 5 
Hence, 
h
0
h
1
=
25
5
= 5 
 
Q2: The maximum speed of a boat in still water is ???? ???? /?? . Now this 
boat is moving downstream in a river flowing at ?? ???? /?? . A man in the 
boat throws a ball vertically upwards with speed of ???? ?? /?? . Range of 
the ball as observed by an observer at rest on the bank is ____ cm. (Take 
?? = ???? ?? /?? ?? ) 
JEE Main 2025 (Online) 29th January Morning Shift 
Answer: 2000 
Solution: 
The first step is to determine the horizontal velocity of the ball as seen by an observer on the bank. 
The velocity of the boat in still water is 27 km/h. Since the boat is moving downstream and the 
river flows at 9 km/h, the actual speed of the boat relative to the bank becomes: 
?? boat 
= 27+ 9 = 36 km/h 
Convert this speed from km/h to m/s by using the conversion factor 1 km/h =
5
18
 m/s : 
?? boat 
= 36 ×
5
18
= 10 m/s 
This is also the horizontal velocity of the ball as seen by the observer on the bank. 
Next, calculate the time the ball spends in the air. The initial vertical velocity of the ball is 10 m/s. 
The time to reach the maximum height ?? up 
 is given by: 
?? up 
=
?? initial 
?? =
10
10
= 1 s 
The total time of flight ?? total 
 is twice the time to reach the maximum height: 
?? total 
= 2× 1 = 2 s 
The horizontal range ?? of the ball is the horizontal velocity multiplied by the total time of flight: 
?? = ?? horizontal 
 × ?? total 
= 10× 2 = 20 m 
Finally, convert this range into centimeters: 
?? = 20× 100= 2000 cm 
Thus, the range of the ball as observed by an observer at rest on the bank is 2000 cm . 
 
Q3: A ball of mass ???????? is projected with velocity ???? ?? /?? at ????
°
 with 
horizontal. The decrease in kinetic energy of the ball during the 
motion from point of projection to highest point is 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 20 J 
B. 5 J 
C. 15 J 
D. zero 
Answer: C 
Solution: 
 
 
 
k
i
=
1
2
mv
2
 
k
f
=
1
2
 m(vcos 60
°
)
2
=
1
8
?? v
2
 
?k = k
i
- k
f
=
3
8
mv
2
=
3
8
× 0.1 × 400= 15 J 
 
Q4: An object of mass ' ?? ' is projected from origin in a vertical xy 
plane at an angle ????
°
 with the x - axis with an initial velocity ?? ?? . The 
magnitude and direction of the angular momentum of the object with 
respect to origin, when it reaches at the maximum height, will be [ ?? is 
acceleration due to gravity] 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 
?? ?? ?? 
3
2v2?? along negative ?? -axis 
B. 
?? ?? ?? 3
2v2?? along positive ?? -axis 
C. 
?? ?? ?? 3
4v2?? along positive ?? -axis 
D. 
?? ?? ?? 3
4v2?? along negative z -axis 
Answer: D 
Solution: 
?? 0
???? =
?? 0
v2
??ˆ +
?? 0
v2
??ˆ. 
At maximum height, the vertical component of the velocity becomes zero. Setting the vertical 
velocity to zero: 
?? 0
v2
- ???? = 0 ? ?? =
?? 0
?? v2
. 
The coordinates at this time are: 
?? =
?? 0
v2
?? =
?? 0
v2
·
?? 0
?? v2
=
?? 0
2
2?? , 
?? =
?? 0
v2
?? -
1
2
?? ?? 2
=
?? 0
2
2?? -
1
2
?? ?? 0
2
2?? 2
=
?? 0
2
2?? -
?? 0
2
4?? =
?? 0
2
4?? . 
Thus, the position vector at maximum height is: 
?? =
?? 0
2
2?? ??ˆ +
?? 0
2
4?? ??ˆ. 
At maximum height, the only nonzero component of velocity is the horizontal one: 
?? =
?? 0
v2
??ˆ. 
The angular momentum about the origin is given by: 
?? ? 
= ?? ?? × ?? . 
Writing out the cross product: 
?? × ?? = (
?? 0
2
2?? ??ˆ +
?? 0
2
4?? ??ˆ) ×
?? 0
v2
??ˆ. 
Since the cross product of ??ˆ with itself is zero and: 
??ˆ × ??ˆ = -?? ˆ
, 
we have: 
?? × ?? =
?? 0
2
4?? ·
?? 0
v2
(-?? ˆ
) = -
?? 0
3
4v2?? ?? ˆ
. 
Thus, the angular momentum is: 
?? ? 
= -
?? ?? 0
3
4v2?? ?? ˆ
. 
This means the magnitude is: 
?? ?? 0
3
4v2?? , 
and its direction is along the negative ?? ˆ
 (or negative ?? -axis). 
 
Q5: The position vector of a moving body at any instant of time is 
given as ?? = (?? ?? ?? ??ˆ - ???? ??ˆ)?? . The magnitude and direction of velocity at 
?? = ?? ?? is, 
JEE Main 2025 (Online) 24th January Evening Shift 
Options: 
A. 5v17 m/s, making an angle of tan
-1
 4 with - ve Y axis 
B. 5v15 m/s, making an angle of tan
-1
 4 with + ve ?? axis 
C. 5v17 m/s, making an angle of tan
-1
 4 with + ve ?? axis 
D. 5v15 m/s, making an angle of tan
-1
 4 with - ve ?? axis 
Answer: A 
Solution: 
r = 5t
2
i ˆ - 5t
ˆ
 
v ? = 10ti ˆ - 5j ˆ 
v ? = 20i ˆ - 5j ˆ  at t = 2sec 
 
 
 
tan ?? =
20
5
= 4 
?? = tan
-1
 4 
From-veY-axis 
 
Q6: Two projectiles are fired with same initial speed from same point 
on ground at angles of ( ????
°
- ?? ) and ( ????
°
+ ?? ), respectively, with the 
horizontal direction. The ratio of their maximum heights attained is : 
JEE Main 2025 (Online) 29th January Morning Shift 
Options: 
A. 
1 + sin ?? 1 - sin ?? 
B. 
1 + sin 2?? 1 - sin 2?? 
C. 
1 - tan ?? 1 + tan ?? 
1 + tan ?? 
D. 
1 - sin 2?? 1 + sin 2?? 
Answer: D 
Solution: 
H
Max 
=
(usin?? )
2
2 g