Vertical Motion

VM-01

A ball is projected vertically upward with a speed of 30 m/s. Taking g = 10 m/s², its velocity after 2 s is

A. 5 m/s upward
B. 10 m/s upward
C. 15 m/s upward
D. 20 m/s upward

Correct Answer: B

Trigger

Velocity after given time

Formula

v = u − gt

Execution

v = 30 − 10(2)

v = 10 m/s upward

Trap: Wrong sign of gravity.

VM-02

A ball is projected vertically upward with a speed of 20 m/s. The height reached after 1 s is

A. 10 m
B. 15 m
C. 20 m
D. 25 m

Correct Answer: B

Trigger

Height after given time

Formula

s = ut − ½gt²

Execution

s = 20(1) − 5(1)²

s = 15 m

Trap: Using velocity equations to find displacement directly.

VM-07

A ball passes a point while moving upward with a speed of 20 m/s. What is its speed when passing the same point while moving downward?

A. 0 m/s
B. 10 m/s
C. 20 m/s
D. 40 m/s

Correct Answer: C

Trigger

Same height recognition

Concept

Vertical Motion Symmetry

Execution

Same position

Same speed magnitude

= 20 m/s

Trap: Using v = u − gt unnecessarily.

VM-08

A stone is dropped from rest. Taking g = 10 m/s², its speed after 4 s is

A. 20 m/s
B. 30 m/s
C. 40 m/s
D. 50 m/s

Correct Answer: C

Trigger

Free fall from rest

Formula

v = gt

Execution

v = 10 × 4

v = 40 m/s

Trap: Using displacement formula instead of velocity.

VM-09

A stone is dropped from a tower and reaches the ground in 3 s. The height of the tower is

A. 30 m
B. 45 m
C. 60 m
D. 90 m

Correct Answer: B

Trigger

Tower drop

Formula

s = ½gt²

Execution

s = 5 × 3²

s = 45 m

Trap: Using v = gt and reporting velocity as height.

VM-10

A ball is thrown vertically downward from a tower with a speed of 10 m/s. Its velocity after 2 s is

A. 10 m/s
B. 20 m/s
C. 30 m/s
D. 40 m/s

Correct Answer: C

Trigger

Downward projection

Formula

v = u + gt

Execution

v = 10 + 10(2)

v = 30 m/s

Trap: Using the sign convention for upward projection.

VM-11

A ball is projected upward from the top of a tower with a speed of 20 m/s. Its velocity at the highest point is

A. 20 m/s upward
B. 20 m/s downward
C. 10 m/s upward
D. 0 m/s

Correct Answer: D

Trigger

Projection from tower

Concept

Highest-Point Condition

v = 0

Trap: Confusing position with state of motion.

VM-12

A freely falling body covers distances in successive seconds in the ratio

A. 1 : 2 : 3 : 4
B. 1 : 3 : 5 : 7
C. 1 : 4 : 9 : 16
D. 2 : 3 : 4 : 5

Correct Answer: B

Trigger

Successive-second motion

Concept

Free-Fall Ratio Rule

Execution

Known Sequence

1 : 3 : 5 : 7 : ...

Trap: Using s ∝ t² directly.

VM-13

A body is dropped from rest. The distance covered during the second second is

A. g/2
B. g
C. 3g/2
D. 2g

Correct Answer: C

Trigger

Distance during nth second

Formula

sₙ = u + (a/2)(2n−1)

Execution

u = 0

n = 2

s₂ = (g/2)(3)

s₂ = 3g/2

Trap: Do not calculate total distance after 2 seconds.

VM-14

A body falls freely for 4 s. The distance covered during the last second is

A. 5 m
B. 15 m
C. 25 m
D. 35 m

Correct Answer: D

Trigger

Last-second motion

Formula

sₙ = u + (a/2)(2n−1)

Execution

u = 0

n = 4

s₄ = (10/2)(7)

s₄ = 35 m

Trap: Do not find total distance after 4 seconds. The question asks only for the last second.

VM-15

Ignoring air resistance, a heavy stone and a light stone are dropped together. Which reaches the ground first?

A. Heavy stone
B. Light stone
C. Both together
D. Cannot be determined

Correct Answer: C

Trigger

Mass dependence question

Concept

Acceleration due to gravity is independent of mass.

Trap: Heavier objects do not fall faster in ideal free fall.

VM-16

A stone is dropped from a tower. Another stone is dropped from the same point 2 seconds later. The separation between them after another 3 seconds is

A. 25 m
B. 50 m
C. 75 m
D. 100 m

Correct Answer: B

Trigger

Delayed release

Formula

s = ½gt²

Execution

First stone:
t = 5 s
s₁ = 5 × 25 = 125 m

Second stone:
t = 3 s
s₂ = 5 × 9 = 45 m

Separation
= 125 − 45
= 80 m

Physics Check: The calculated separation is 80 m. None of the options match. Question or options should be corrected before publishing.

VM-17

Balls are projected vertically upward at intervals of 2 seconds. For successive balls to meet, the minimum launch speed must be

A. g/2
B. g
C. Greater than g
D. Greater than 2g

Correct Answer: C

Concept

Time of Flight

T = 2u/g

Execution

For meeting, the first ball must remain in air for more than 2 s.

2u/g > 2

u > g

Trap: Using maximum height instead of time of flight.

VM-18

Taking upward as positive, the value of acceleration due to gravity is

A. +g
B. −g
C. 0
D. Depends on velocity

Correct Answer: B

Trigger

Sign Convention

Concept

Gravity always acts downward.

Execution

Upward positive

Gravity downward

a = −g

Trap: Sign depends on chosen coordinate system, not on direction of motion.

VM-19

A ball is projected vertically upward with a speed of 40 m/s. Taking g = 10 m/s², its total time of flight is

A. 4 s
B. 6 s
C. 8 s
D. 10 s

Correct Answer: C

Trigger

Time of Flight

Formula

T = 2u/g

Execution

T = 2(40)/10

T = 8 s

Trap: Do not confuse ascent time with total time of flight.

VM-20

A ball is projected vertically upward. At half of its maximum height, its acceleration is

A. Zero
B. g/2
C. g upward
D. g downward

Correct Answer: D

Concept

Gravity remains constant throughout the motion.

Execution

Acceleration

= g downward

Trap: Acceleration does not decrease with height for standard near-Earth problems.

VM-21

At the highest point of vertical motion, which quantity becomes zero?

A. Velocity only
B. Acceleration only
C. Velocity and acceleration
D. Gravity

Correct Answer: A

Concept

Highest Point State

Execution

Velocity = 0

Acceleration = g downward

Trap: Many students incorrectly assume acceleration becomes zero.

VM-22

A body is projected vertically upward. Which graph correctly represents its acceleration versus time?

A. Increasing straight line
B. Decreasing straight line
C. Horizontal line at +g
D. Horizontal line at −g

Correct Answer: D

Trigger

Graph Recognition

Concept

Acceleration remains constant.

Execution

With upward positive,

a = −g

constant

Trap: Do not draw acceleration changing with time.

VM-23

A ball is projected upward. Which quantity continuously decreases during the upward journey?

A. Acceleration
B. Speed
C. Gravity
D. Mass

Correct Answer: B

Concept

Gravity opposes upward motion.

Execution

Speed decreases uniformly until the highest point.

Trap: Acceleration remains constant. Only speed changes.

VM-24

A body is dropped from rest. Its velocity-time graph is

A. Horizontal line
B. Straight line through origin
C. Parabola
D. Hyperbola

Correct Answer: B

Trigger

Graph Intelligence

Formula

v = gt

Execution

Velocity proportional to time

Straight line through origin

Trap: Do not confuse v-t and s-t graphs.

VM-25

For a body moving only under gravity, the acceleration-time graph is

A. Horizontal line
B. Vertical line
C. Parabola
D. Curve opening upward

Correct Answer: A

Concept

Gravity provides constant acceleration.

Execution

a = constant

Therefore

a-t graph is horizontal.

Trap: Constant acceleration means horizontal graph, not sloping graph.

VM-03

A stone is projected vertically upward with a speed of 40 m/s. The time taken to reach the highest point is

A. 2 s
B. 3 s
C. 4 s
D. 5 s

Correct Answer: C

Trigger

Highest point

Formula

v = u − gt

Execution

0 = 40 − 10t

t = 4 s

Trap: Using total time of flight instead of ascent time.

VM-04

A ball is projected vertically upward with a speed of 20 m/s. The maximum height reached is

A. 10 m
B. 15 m
C. 20 m
D. 25 m

Correct Answer: C

Trigger

Maximum height

Formula

v² = u² − 2gh

Execution

0 = 20² − 20h

h = 20 m

Trap: Using time-based equations unnecessarily.

VM-05

At the highest point of its motion, a vertically projected ball has

A. Velocity zero and acceleration zero
B. Velocity upward and acceleration downward
C. Velocity zero and acceleration downward
D. Velocity downward and acceleration zero

Correct Answer: C

Trigger

Highest point state

Key Concept

At highest point:

v = 0

a = g downward

Trap: v = 0 does not imply a = 0. Gravity acts throughout the motion.

VM-06

A ball is projected vertically upward with a speed of 30 m/s. Its speed when it returns to the point of projection is

A. 0 m/s
B. 15 m/s
C. 30 m/s
D. 60 m/s

Correct Answer: C

Trigger

Return to launch point

Concept

Vertical Motion Symmetry

Execution

Returns to the same height

Speed magnitude remains unchanged

= 30 m/s

Trap: Assuming speed keeps decreasing throughout the motion.

VM-07

A ball passes a point while moving upward with a speed of 20 m/s. What is its speed when passing the same point while moving downward?

A. 0 m/s
B. 10 m/s
C. 20 m/s
D. 40 m/s

Correct Answer: C

Trigger

Same height recognition

Concept

Vertical Motion Symmetry

Execution

Same position

Same speed magnitude

= 20 m/s

Trap: Using v = u − gt unnecessarily.

VM-08

A stone is dropped from rest. Taking g = 10 m/s², its speed after 4 s is

A. 20 m/s
B. 30 m/s
C. 40 m/s
D. 50 m/s

Correct Answer: C

Trigger

Free fall from rest

Formula

v = gt

Execution

v = 10 × 4

v = 40 m/s

Trap: Using displacement formula instead of velocity.

VM-09

A stone is dropped from a tower and reaches the ground in 3 s. The height of the tower is

A. 30 m
B. 45 m
C. 60 m
D. 90 m

Correct Answer: B

Trigger

Tower drop

Formula

s = ½gt²

Execution

s = 5 × 3²

s = 45 m

Trap: Using v = gt and reporting velocity as height.

VM-10

A ball is thrown vertically downward from a tower with a speed of 10 m/s. Its velocity after 2 s is

A. 10 m/s
B. 20 m/s
C. 30 m/s
D. 40 m/s

Correct Answer: C

Trigger

Downward projection

Formula

v = u + gt

Execution

v = 10 + 10(2)

v = 30 m/s

Trap: Using the sign convention for upward projection.

VM-11

A ball is projected upward from the top of a tower with a speed of 20 m/s. Its velocity at the highest point is

A. 20 m/s upward
B. 20 m/s downward
C. 10 m/s upward
D. 0 m/s

Correct Answer: D

Trigger

Projection from tower

Concept

Highest-Point Condition

v = 0

Trap: Confusing position with state of motion.

VM-12

A freely falling body covers distances in successive seconds in the ratio

A. 1 : 2 : 3 : 4
B. 1 : 3 : 5 : 7
C. 1 : 4 : 9 : 16
D. 2 : 3 : 4 : 5

Correct Answer: B

Trigger

Successive-second motion

Concept

Free-Fall Ratio Rule

Execution

Known Sequence

1 : 3 : 5 : 7 : ...

Trap: Using s ∝ t² directly.

VM-13

A body is dropped from rest. The distance covered during the second second is

A. g/2
B. g
C. 3g/2
D. 2g

Correct Answer: C

Trigger

Distance during nth second

Formula

sₙ = u + (a/2)(2n−1)

Execution

u = 0

n = 2

s₂ = (g/2)(3)

s₂ = 3g/2

Trap: Do not calculate total distance after 2 seconds.

VM-14

A body falls freely for 4 s. The distance covered during the last second is

A. 5 m
B. 15 m
C. 25 m
D. 35 m

Correct Answer: D

Trigger

Last-second motion

Formula

sₙ = u + (a/2)(2n−1)

Execution

u = 0

n = 4

s₄ = (10/2)(7)

s₄ = 35 m

Trap: Do not find total distance after 4 seconds. The question asks only for the last second.

VM-15

Ignoring air resistance, a heavy stone and a light stone are dropped together. Which reaches the ground first?

A. Heavy stone
B. Light stone
C. Both together
D. Cannot be determined

Correct Answer: C

Trigger

Mass dependence question

Concept

Acceleration due to gravity is independent of mass.

Trap: Heavier objects do not fall faster in ideal free fall.

VM-16

A stone is dropped from a tower. Another stone is dropped from the same point 2 seconds later. The separation between them after another 3 seconds is

A. 25 m
B. 50 m
C. 75 m
D. 100 m

Correct Answer:C

Trigger

Delayed release

Formula

s = ½gt²

Execution

First stone:
t = 5 s
s₁ = 5 × 25 = 125 m

Second stone:
t = 3 s
s₂ = 5 × 9 = 45 m

Separation
= 125 − 45
= 80 m

Physics Check: The calculated separation is 80 m. None of the options match. Question or options should be corrected before publishing.

VM-17

Balls are projected vertically upward at intervals of 2 seconds. For successive balls to meet, the minimum launch speed must be

A. g/2
B. g
C. Greater than g
D. Greater than 2g

Correct Answer: C

Concept

Time of Flight

T = 2u/g

Execution

For meeting, the first ball must remain in air for more than 2 s.

2u/g > 2

u > g

Trap: Using maximum height instead of time of flight.

VM-18

Taking upward as positive, the value of acceleration due to gravity is

A. +g
B. −g
C. 0
D. Depends on velocity

Correct Answer: B

Trigger

Sign Convention

Concept

Gravity always acts downward.

Execution

Upward positive

Gravity downward

a = −g

Trap: Sign depends on chosen coordinate system, not on direction of motion.

VM-19

A ball is projected vertically upward with a speed of 40 m/s. Taking g = 10 m/s², its total time of flight is

A. 4 s
B. 6 s
C. 8 s
D. 10 s

Correct Answer: C

Trigger

Time of Flight

Formula

T = 2u/g

Execution

T = 2(40)/10

T = 8 s

Trap: Do not confuse ascent time with total time of flight.

VM-20

A ball is projected vertically upward. At half of its maximum height, its acceleration is

A. Zero
B. g/2
C. g upward
D. g downward

Correct Answer: D

Concept

Gravity remains constant throughout the motion.

Execution

Acceleration

= g downward

Trap: Acceleration does not decrease with height for standard near-Earth problems.

VM-21

At the highest point of vertical motion, which quantity becomes zero?

A. Velocity only
B. Acceleration only
C. Velocity and acceleration
D. Gravity

Correct Answer: A

Concept

Highest Point State

Execution

Velocity = 0

Acceleration = g downward

Trap: Many students incorrectly assume acceleration becomes zero.

VM-22

A body is projected vertically upward. Which graph correctly represents its acceleration versus time?

A. Increasing straight line
B. Decreasing straight line
C. Horizontal line at +g
D. Horizontal line at −g

Correct Answer: D

Trigger

Graph Recognition

Concept

Acceleration remains constant.

Execution

With upward positive,

a = −g

constant

Trap: Do not draw acceleration changing with time.

VM-23

A ball is projected upward. Which quantity continuously decreases during the upward journey?

A. Acceleration
B. Speed
C. Gravity
D. Mass

Correct Answer: B

Concept

Gravity opposes upward motion.

Execution

Speed decreases uniformly until the highest point.

Trap: Acceleration remains constant. Only speed changes.

VM-24

A body is dropped from rest. Its velocity-time graph is

A. Horizontal line
B. Straight line through origin
C. Parabola
D. Hyperbola

Correct Answer: B

Trigger

Graph Intelligence

Formula

v = gt

Execution

Velocity proportional to time

Straight line through origin

Trap: Do not confuse v-t and s-t graphs.

VM-25

For a body moving only under gravity, the acceleration-time graph is

A. Horizontal line
B. Vertical line
C. Parabola
D. Curve opening upward

Correct Answer: A

Concept

Gravity provides constant acceleration.

Execution

a = constant

Therefore

a-t graph is horizontal.

Trap: Constant acceleration means horizontal graph, not sloping graph.

VM-01

VM-02

VM-07

VM-08

VM-09

VM-10

VM-11

VM-12

VM-13

VM-14

VM-15

VM-16

VM-17

VM-18

VM-19

VM-20

VM-21

VM-22

VM-23

VM-24

VM-25

Continue Learning

Uniformly Accelerated Motion

Relative Motion

Pursuit Problems

Revision Handbook

VM-03

VM-04

VM-05

VM-06

VM-07

VM-08

VM-09

VM-10

VM-11

VM-12

VM-13

VM-14

VM-15

VM-16

VM-17

VM-18

VM-19

VM-20

VM-21

VM-22

VM-23

VM-24

VM-25

Continue Learning

Uniformly Accelerated Motion

Relative Motion

Pursuit Problems

Revision Handbook