Motion in a Plane: JEE Main Previous Year Questions (2021-2026)

 Page 1


JEE Main Previous Year Questions (2021-2026): 
Motion in a Plane  
 
(January 2026) 
 
 
Q1: A boy throws a ball into air at 45
°
 from the horizontal to land it on a roof of a building 
of height H. If the ball attains maximum height in 2 s and lands on the building in 3 s after 
launch, then value of H is _____ m. 
(g = 10 m/s
2
) 
(a) 20 
(b) 25 
(c) 10 
(d) 15 
Ans: (d) 
Sol: Let u be the initial velocity and ? = 45
°
 be the angle of projection. 
 
The initial vertical velocity is u
y
 = u sin ? 
The ball attains maximum height in t
max
 = 2 s. 
At the maximum height, the vertical component of the velocity becomes zero (vy = 0).  
Using the first equation of motion (v = u + at) :  
v
y
 = u
y
 - gt  
? 0 = u
y
 - (10) (2) ? u
y
 = 20 m/s  
So, the initial vertical velocity is 20 m/s.  
Page 2


JEE Main Previous Year Questions (2021-2026): 
Motion in a Plane  
 
(January 2026) 
 
 
Q1: A boy throws a ball into air at 45
°
 from the horizontal to land it on a roof of a building 
of height H. If the ball attains maximum height in 2 s and lands on the building in 3 s after 
launch, then value of H is _____ m. 
(g = 10 m/s
2
) 
(a) 20 
(b) 25 
(c) 10 
(d) 15 
Ans: (d) 
Sol: Let u be the initial velocity and ? = 45
°
 be the angle of projection. 
 
The initial vertical velocity is u
y
 = u sin ? 
The ball attains maximum height in t
max
 = 2 s. 
At the maximum height, the vertical component of the velocity becomes zero (vy = 0).  
Using the first equation of motion (v = u + at) :  
v
y
 = u
y
 - gt  
? 0 = u
y
 - (10) (2) ? u
y
 = 20 m/s  
So, the initial vertical velocity is 20 m/s.  
The ball lands on the roof of the building at t = 3s.  
This means at t = 3, the vertical displacement (y) of the ball is equal to the height of the building 
(H).  
Using the second equation of motion  
 
Substituting the given values : 
1. y = H (Height of the building)  
2. u
y
 = 20 m / s (Calculated in Step 2)  
3. t = 3 s (Given time of landing) 
 
Therefore, the height of the building is 15 m. Hence, the correct option is (D). 
 
Q2: A projectile is thrown upward at an angle 60
°
 with the horizontal. The speed of the 
projectile is 20 m / s when its direction of motion is 45
°
 with the horizontal.  
The initial speed of the projectile is _____ m/s.  
(a) 20v3  
(b) 20v2  
(c) 40  
(d) 40v2 
Ans: (b) 
Sol: 
When a projectile is launched into the air, it moves in two dimensions: horizontally (x - axis) and 
vertically (y - axis).  
Along vertical direction gravity constantly pulls the projectile downward. Therefore, there is a 
constant downward acceleration (a
y
 = - g).  
Along horizontal direction there are no horizontal forces (assuming no air resistance). According 
to Newton's First Law, an object in motion stays in motion with the same speed and in the same 
direction unless acted upon by an unbalanced force. Therefore, the horizontal acceleration is 
zero (a
x
 = 0). Because a
x
 = 0 , the horizontal component of the projectile's velocity remains 
perfectly constant throughout its entire journey. 
Page 3


JEE Main Previous Year Questions (2021-2026): 
Motion in a Plane  
 
(January 2026) 
 
 
Q1: A boy throws a ball into air at 45
°
 from the horizontal to land it on a roof of a building 
of height H. If the ball attains maximum height in 2 s and lands on the building in 3 s after 
launch, then value of H is _____ m. 
(g = 10 m/s
2
) 
(a) 20 
(b) 25 
(c) 10 
(d) 15 
Ans: (d) 
Sol: Let u be the initial velocity and ? = 45
°
 be the angle of projection. 
 
The initial vertical velocity is u
y
 = u sin ? 
The ball attains maximum height in t
max
 = 2 s. 
At the maximum height, the vertical component of the velocity becomes zero (vy = 0).  
Using the first equation of motion (v = u + at) :  
v
y
 = u
y
 - gt  
? 0 = u
y
 - (10) (2) ? u
y
 = 20 m/s  
So, the initial vertical velocity is 20 m/s.  
The ball lands on the roof of the building at t = 3s.  
This means at t = 3, the vertical displacement (y) of the ball is equal to the height of the building 
(H).  
Using the second equation of motion  
 
Substituting the given values : 
1. y = H (Height of the building)  
2. u
y
 = 20 m / s (Calculated in Step 2)  
3. t = 3 s (Given time of landing) 
 
Therefore, the height of the building is 15 m. Hence, the correct option is (D). 
 
Q2: A projectile is thrown upward at an angle 60
°
 with the horizontal. The speed of the 
projectile is 20 m / s when its direction of motion is 45
°
 with the horizontal.  
The initial speed of the projectile is _____ m/s.  
(a) 20v3  
(b) 20v2  
(c) 40  
(d) 40v2 
Ans: (b) 
Sol: 
When a projectile is launched into the air, it moves in two dimensions: horizontally (x - axis) and 
vertically (y - axis).  
Along vertical direction gravity constantly pulls the projectile downward. Therefore, there is a 
constant downward acceleration (a
y
 = - g).  
Along horizontal direction there are no horizontal forces (assuming no air resistance). According 
to Newton's First Law, an object in motion stays in motion with the same speed and in the same 
direction unless acted upon by an unbalanced force. Therefore, the horizontal acceleration is 
zero (a
x
 = 0). Because a
x
 = 0 , the horizontal component of the projectile's velocity remains 
perfectly constant throughout its entire journey. 
  
If a projectile is launched with an initial speed u at an angle ?, its horizontal component is  
u
x
 = u cos ?  
At any later point in time, if the projectile has a new speed v and is moving at a new angle a, its 
horizontal component is :  
v
x
 = v cos a  
Since the horizontal velocity never changes, these two expressions must be equal :  
u
x
 = v
x
  
? u cos   ? = v cos   a  
Substituting the given values: 
 
Therefore, the initial speed of projectile is 20v2 m/s.  
Hence, the correct option is (b). 
 
Q3: A river of width 200 m is flowing from west to east with a speed of 18 km/h. A boat, 
moving with speed of 36 km/h in still water, is made to travel one-round trip (bank to bank 
of the river). Minimum time taken by the boat for this journey and also the displacement 
Page 4


JEE Main Previous Year Questions (2021-2026): 
Motion in a Plane  
 
(January 2026) 
 
 
Q1: A boy throws a ball into air at 45
°
 from the horizontal to land it on a roof of a building 
of height H. If the ball attains maximum height in 2 s and lands on the building in 3 s after 
launch, then value of H is _____ m. 
(g = 10 m/s
2
) 
(a) 20 
(b) 25 
(c) 10 
(d) 15 
Ans: (d) 
Sol: Let u be the initial velocity and ? = 45
°
 be the angle of projection. 
 
The initial vertical velocity is u
y
 = u sin ? 
The ball attains maximum height in t
max
 = 2 s. 
At the maximum height, the vertical component of the velocity becomes zero (vy = 0).  
Using the first equation of motion (v = u + at) :  
v
y
 = u
y
 - gt  
? 0 = u
y
 - (10) (2) ? u
y
 = 20 m/s  
So, the initial vertical velocity is 20 m/s.  
The ball lands on the roof of the building at t = 3s.  
This means at t = 3, the vertical displacement (y) of the ball is equal to the height of the building 
(H).  
Using the second equation of motion  
 
Substituting the given values : 
1. y = H (Height of the building)  
2. u
y
 = 20 m / s (Calculated in Step 2)  
3. t = 3 s (Given time of landing) 
 
Therefore, the height of the building is 15 m. Hence, the correct option is (D). 
 
Q2: A projectile is thrown upward at an angle 60
°
 with the horizontal. The speed of the 
projectile is 20 m / s when its direction of motion is 45
°
 with the horizontal.  
The initial speed of the projectile is _____ m/s.  
(a) 20v3  
(b) 20v2  
(c) 40  
(d) 40v2 
Ans: (b) 
Sol: 
When a projectile is launched into the air, it moves in two dimensions: horizontally (x - axis) and 
vertically (y - axis).  
Along vertical direction gravity constantly pulls the projectile downward. Therefore, there is a 
constant downward acceleration (a
y
 = - g).  
Along horizontal direction there are no horizontal forces (assuming no air resistance). According 
to Newton's First Law, an object in motion stays in motion with the same speed and in the same 
direction unless acted upon by an unbalanced force. Therefore, the horizontal acceleration is 
zero (a
x
 = 0). Because a
x
 = 0 , the horizontal component of the projectile's velocity remains 
perfectly constant throughout its entire journey. 
  
If a projectile is launched with an initial speed u at an angle ?, its horizontal component is  
u
x
 = u cos ?  
At any later point in time, if the projectile has a new speed v and is moving at a new angle a, its 
horizontal component is :  
v
x
 = v cos a  
Since the horizontal velocity never changes, these two expressions must be equal :  
u
x
 = v
x
  
? u cos   ? = v cos   a  
Substituting the given values: 
 
Therefore, the initial speed of projectile is 20v2 m/s.  
Hence, the correct option is (b). 
 
Q3: A river of width 200 m is flowing from west to east with a speed of 18 km/h. A boat, 
moving with speed of 36 km/h in still water, is made to travel one-round trip (bank to bank 
of the river). Minimum time taken by the boat for this journey and also the displacement 
along the river bank are ______ and ______ respectively. 
(a) 20 s and 100 m 
(b) 40 s and 100 m 
(c) 40 s and 200 m 
(d) 40 s and 0 m 
Ans: (c)  
Sol:  
The speed of river is  
The speed of boat in still water is  
The width of river is w = 200 m  
 
The time taken to cross a river is determined solely by the component of the boat's velocity 
perpendicular to the river banks. 
 
Where theta is the angle with the bank. Time is minimum when the boat heads straight across 
(? = 90
°
), so sin   ? = 1. 
 
Also, while returning to initial bank in minimum time, 
Page 5


JEE Main Previous Year Questions (2021-2026): 
Motion in a Plane  
 
(January 2026) 
 
 
Q1: A boy throws a ball into air at 45
°
 from the horizontal to land it on a roof of a building 
of height H. If the ball attains maximum height in 2 s and lands on the building in 3 s after 
launch, then value of H is _____ m. 
(g = 10 m/s
2
) 
(a) 20 
(b) 25 
(c) 10 
(d) 15 
Ans: (d) 
Sol: Let u be the initial velocity and ? = 45
°
 be the angle of projection. 
 
The initial vertical velocity is u
y
 = u sin ? 
The ball attains maximum height in t
max
 = 2 s. 
At the maximum height, the vertical component of the velocity becomes zero (vy = 0).  
Using the first equation of motion (v = u + at) :  
v
y
 = u
y
 - gt  
? 0 = u
y
 - (10) (2) ? u
y
 = 20 m/s  
So, the initial vertical velocity is 20 m/s.  
The ball lands on the roof of the building at t = 3s.  
This means at t = 3, the vertical displacement (y) of the ball is equal to the height of the building 
(H).  
Using the second equation of motion  
 
Substituting the given values : 
1. y = H (Height of the building)  
2. u
y
 = 20 m / s (Calculated in Step 2)  
3. t = 3 s (Given time of landing) 
 
Therefore, the height of the building is 15 m. Hence, the correct option is (D). 
 
Q2: A projectile is thrown upward at an angle 60
°
 with the horizontal. The speed of the 
projectile is 20 m / s when its direction of motion is 45
°
 with the horizontal.  
The initial speed of the projectile is _____ m/s.  
(a) 20v3  
(b) 20v2  
(c) 40  
(d) 40v2 
Ans: (b) 
Sol: 
When a projectile is launched into the air, it moves in two dimensions: horizontally (x - axis) and 
vertically (y - axis).  
Along vertical direction gravity constantly pulls the projectile downward. Therefore, there is a 
constant downward acceleration (a
y
 = - g).  
Along horizontal direction there are no horizontal forces (assuming no air resistance). According 
to Newton's First Law, an object in motion stays in motion with the same speed and in the same 
direction unless acted upon by an unbalanced force. Therefore, the horizontal acceleration is 
zero (a
x
 = 0). Because a
x
 = 0 , the horizontal component of the projectile's velocity remains 
perfectly constant throughout its entire journey. 
  
If a projectile is launched with an initial speed u at an angle ?, its horizontal component is  
u
x
 = u cos ?  
At any later point in time, if the projectile has a new speed v and is moving at a new angle a, its 
horizontal component is :  
v
x
 = v cos a  
Since the horizontal velocity never changes, these two expressions must be equal :  
u
x
 = v
x
  
? u cos   ? = v cos   a  
Substituting the given values: 
 
Therefore, the initial speed of projectile is 20v2 m/s.  
Hence, the correct option is (b). 
 
Q3: A river of width 200 m is flowing from west to east with a speed of 18 km/h. A boat, 
moving with speed of 36 km/h in still water, is made to travel one-round trip (bank to bank 
of the river). Minimum time taken by the boat for this journey and also the displacement 
along the river bank are ______ and ______ respectively. 
(a) 20 s and 100 m 
(b) 40 s and 100 m 
(c) 40 s and 200 m 
(d) 40 s and 0 m 
Ans: (c)  
Sol:  
The speed of river is  
The speed of boat in still water is  
The width of river is w = 200 m  
 
The time taken to cross a river is determined solely by the component of the boat's velocity 
perpendicular to the river banks. 
 
Where theta is the angle with the bank. Time is minimum when the boat heads straight across 
(? = 90
°
), so sin   ? = 1. 
 
Also, while returning to initial bank in minimum time, 
 
Total time of journey is, 
 
When the boat heads straight across to minimize time, the river flow carries it downstream. This 
horizontal distance is, Drift (one way) min = v
r
 × t
min
 
1 ? 2
 = 5 × 20 s = 100 m east 
The boat again heads straight across to minimize time. The river still flows from west to east, so 
the boat drifts another 100 m east. 
Total Displacement along the bank: 
?x = 100 m (east) + 100 m (east) = 200 m  
Therefore, the minimum time is 40 s and the displacement along the bank is 200 m . Hence, the 
correct option is (C).